Transitivity of convergence of sequence in $L^1$ and $BV$

Question

Let $(f_n)_n\in BV([0,T])$ a sequence of functions of bounded variation converging to some $f\in BV([0,T])$, that is to say

$$\lim_{n\rightarrow+\infty}\parallel f_n-f\parallel_{BV([0,T])} = 0.$$ and such that $$\forall n, f_n(0)=f(0)$$ Let $(g_n)_n\in BV([0,T])$ a sequence of function of bounded variations such that the following holds $$\lim_{n\rightarrow +\infty}\parallel g_n-f_n\parallel_{L^1([0,T])} = 0$$

Does the following holds ?

$$\lim_{n\rightarrow +\infty}\parallel g_n-f\parallel_{L^1([0,T])}=0$$

If not, is there a topology where some convergence result holds ?

Answer 1

Counter-example: $f_n(x)=g_n(x)=n$ and $f(x)=0$.

Answer for the edited version: Convefence in BV norm together with $f_n(0)=f(0)=0$ impiies uniform convergence, hence also convergence in $L^{1}$. [$|h(x)-h(0)| \leq \|h\|_{BV}$ if $h(0)=0$]. Hence, the answer is YES.