Pointwise Convergent Composition Operator in BV I am only considering functions of bounded variation in the classical sense, i.e. functions $f:[0,1]\to\mathbb R$, s.t. $||f||_{BV}<\infty$, where
\begin{align*}
||f||_{BV}=||f||_\infty+\sup\sum_{j=1}^n|f(t_j)-f(t_{j-1})|,
\end{align*}
and the supremum is taken over all partitions $0=t_0<\ldots<t_n=1$ of $[0,1]$. The set of all such functions is denoted by $BV$, and the subset $BV_0$ is defined as the set of all functions $x\in BV$ such that $x(t)\in[0,1]$ for all $t\in[0,1]$.
One can show that $||\cdot||_{BV}$ defines a norm on $BV$ and makes $BV$ a complete vector space.
My question now is as follows.

Let $f_n:[0,1]\to\mathbb R$ be a sequence of functions satisfying $|f_n(u)-f_n(v)|\leq|u-v|$ for all $u,v\in[0,1]$, $n\in\mathbb N$, and $||f_n||_{BV}\to0$ as $n\to\infty$.
(1) Is it always true that $||f_n\circ x||_{BV}\to0$ as $n\to\infty$ for all $x\in BV_0$? (2) If not, is (1) true at least for a subsequence of $(f_n)$?

Note that $||f_n||_{BV}\to 0$ implies that $(f_n)$ converges uniformly to 0.
If the Lipschitz constants $L_n$ of $f_n$ converge to 0, then the answer is clearly yes, because $||f_n\circ x||_{BV}\leq ||f_n||_\infty+L_n||x||_{BV}$. But what happens in the general case? I only know that $L_n\leq 1$ for all $n\in\mathbb N$.
Any help is highly appreciated. Thank you in advance!
 A: OK, here goes.
Lemma: Let $x\in BV,\varepsilon>0$. There exists $\delta=\delta(x,\varepsilon)$ such that for any system $\mathcal I$ of pairwise disjoint open intervals $I_k$ (which may be finite or infinite) of total length $\sum_k|I_k|< \delta$ and for any $t_1<t_2<\dots<t_m$, we have
$$
{\sum_j}^{(\mathcal I)}|x(t_j)-x(t_{j-1})|\le \varepsilon
$$
where ${\sum_j}^{(\mathcal I)}$ denotes the sum taken over such indices $j$ that $x(t_{j-1}),x(t_j)\in I_k$ for some $k$ (we'll call the corresponding jumps "internal" and the other ones "external").
Proof: Choose $0=s_1<s_2<\dots<s_N=1$ so that 
$$
\sum_m|x(s_m)-x(s_{m-1})|\ge \operatorname{Var}x-\frac\varepsilon 2
$$
Put $\delta=\frac{\varepsilon}{4N}$. Consider the partition determined by $s_m$ and $t_j$ together. Notice that for each $m$, the external jumps between $x(t_j)$ with $t_j$ lying in the interval $[s_{m-1},s_m]$ together with the "endpoint" jumps on that interval should cover $[x(s_{m-1}),x(s_m)]\setminus(\cup_k I_k)$, so their total length is at least 
$|x(s_m)-x(s_{m-1})|-\delta$ and their total contribution to the variation is thus $\ge \sum_m|x(s_m)-x(s_{m-1})|-N\delta$. As to the internal jumps, only 
$N$ of them can be split by $s_m$ and each of them is at most $\delta$. Thus, their contribution is at least 
${\sum_j}^{(\mathcal I)}|x(t_j)-x(t_{j-1})|-N\delta$. Since together they  cannot contribute more than $\operatorname{Var}x$, the lemma follows.
Now take an $L$-Lipschitz function $f$ with $\|f\|_{BV}<\frac 14\varepsilon\delta$. Consider all intervals $[s,t]$ with $|f(t)-f(s)|>\varepsilon(t-s)$. Notice that the total length of any disjoint family of such intervals is at most $\varepsilon^{-1}\|f\|_{BV}<\frac 14\delta$, so, by Vitali covering lemma, we can find an open set $U\subset \mathbb R$ of length $|U|<\delta$ such that every such interval is contained in $U$. Let $\mathcal I$ be the family of all constituting intervals of $U$. Then, by the lemma, the sum of the extensions by $f$ of internal jumps $|f(x(t_j))-f(x(t_{j-1}))|\le L|x(t_j)-x(t_{j-1})|$ is at most $L\varepsilon$ while the sum of the extensions of external jumps is at most $\varepsilon\operatorname{Var}x$ (because  external jumps see $f$ as $\varepsilon$-Lipschitz rather than $L$-Lipschitz).
That's it. Feel free to ask questions if something is unclear.
