Limiting total variation attached to sequence of uniformly vanishing functions of bounded variation

Let $$(f_n)_n$$ be a sequence of real functions of a single real variable with compact support in $$[0,1]$$ and of bounded variation all of them. Let the sequence be uniformly convergent to $$0$$. Is it true that the sequence of their total variations (on $$[0,1]$$), say $$(V_0^1(f_n))_n$$, is convergent to $$0$$?

Definitely not. Consider the sequence $$f_n(x)=\frac{1}{n}\cos(2n\pi x)$$. This converges uniformly to $$0$$ on $$[0,1]$$, but $$V_0^1(f_n)>1$$ for all $$n$$.
• I wonder, would it make a difference if $f_n = F_n - F$, where $F_n$ and $F$ are cumulative distribution functions such that $(F_n)_n$ is uniformly convergent to $F$? This is more tightly related to my original problem. – Marcos Dec 29 '18 at 6:30
• I can't think of this on the top of my head, but I imagine you'll have better results. I'll come back to this after I've had some sleep. Are there any other hypotheses on the $F_n$ and $F$ which would simplify this? – Aweygan Dec 29 '18 at 7:05
• Thank you very much indeed for your effort. I am not sure that this helps, but every $F_n$ is an increasing step function with precisely $n+1$ jumps uniformly spaced on $[0,1]$ at $x_k = k/n$ for $k=0, \dotsc, n$ (but not in general of the same height all of them), and it is bounded (as a cdf) by $0$ from below and by $1$ from above. – Marcos Dec 29 '18 at 7:16