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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 $0$?

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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 $n$.

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  • $\begingroup$ 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. $\endgroup$ – Marcos Dec 29 '18 at 6:30
  • $\begingroup$ 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? $\endgroup$ – Aweygan Dec 29 '18 at 7:05
  • $\begingroup$ 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. $\endgroup$ – Marcos Dec 29 '18 at 7:16

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