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Suppose $f,g:[0,1]\rightarrow[0,1]$, $g>0$, and $f/g$ is of bounded variation. If $f_n, f \in C^1[0,1]$ and $f_n \rightarrow f$ in $C^1$. Does it follow that $\exists N$ such that $f_n/g$ is of uniform variation for $n>N$?

Here by "$f_n\to f$ in $C^1$" we mean that $f_n\to f$ and $f'_n\f'$ uniformly on the unit interval.

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Are there any hypotheses missing? What if $1/g$ is not of bounded variation, $f=0$, and $f_n=\frac{1}{n}$? – Jonas Meyer Apr 27 '11 at 14:56
The answer to your first question is yes. I was looking for minimal conditions on $f_n$ and $g$ that imply uniformly bounded variation. your counterexample shows other assumptions are needed. – Peyman Apr 27 '11 at 19:20
What if $g \in C^1$? Does the following proof work: let $h_n=f_n/g$, $h=f/g$. Then $V(h_n)-V(h) \leq \int |h_n'-h'| \rightarrow 0$. Therefore $\{V(h_n)\}$ is uniformly bounded. $V()$ denotes the total variation of a function. – Peyman Apr 27 '11 at 19:24
Yes, I think so. – Jonas Meyer Apr 27 '11 at 22:18

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