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Let $f(t,v)$, $t\in[0,T]$, $v\in R$ be absolutely continuous in $t$ for every fixed $v$ and absolutely continuous in $v$ for every fixed $t$. Let $X(t)$, $t\in [0,T]$ be a continuous function of unbounded variation on every interval $[t_1,t_2]\subset[0,T]$. I propose that if for every $t$ $f'_v(t,v)\neq0$ a.e. then $f(t,X(t))$ in a function of unbounded variation.

Need help in proving this. I'll be thankful for any textbook recommendations on related topics.

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Clearly false as stated. Take $f(t,v)=v^2$. Then you are claiming that $X^2$ is of unbounded variation if $X$ is of unbounded variation on every interval. Now take $X$ to be $+1$ at rationals and $-1$ at irrationals. –  fedja Dec 16 '11 at 1:17
    
Sorry, my fault. I forgot to mention that $X(t)$ is a continuous function. –  niyazets Dec 16 '11 at 2:07
    
That won't save you. Take any continuous function $X(t)$ with modulus of continuity $\omega(h)=o(\sqrt h)$. Let $f(t,v)$ be the distance from $(t,v)$ to the graph of $X$ in $\mathbb R^2$. $f$ is Lipschitz and for fixed $t$, $f_v(t,v)\ne 0$ for almost all $v\ne X(t)$ (just think where the closest point can possibly be and how the graph can go from there). However $f(t,X(t))$ is identically $0$. –  fedja Dec 16 '11 at 14:40
    
If I got you right, then $f(t,v) = \min_s \sqrt{(X(s) - v)^2 + (t-s)^2}$ which has unbounded variation for every fixed $v$, while I consider functions with bounded variation. –  niyazets Dec 17 '11 at 7:52
    
And why are you so sure that it has "unbounded variation for every fixed $v$"? I'm sure that the variation is bounded because distance to anything is Lipschitz. What is your argument? –  fedja Dec 18 '11 at 14:29
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