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f is a absolutely continuous function on $[a,b]$,let $V_a^x$ denote the total variation of [a,x] prove that $$①\int_a^b\vert f'(t) \vert dt=V_{a}^b $$ 1. How to prove $\frac{d}{dx}V_a^x= \vert f'(x) \vert$ $a.e.$ I might prove $f,V_a^x$ are both absolutely continuous, and $\vert f'(x) \vert \le \frac{d}{dx}V_a^x$, but I can't prove the reverse inequality.

2.Without the first conclusion,let$$p(x)=\frac{1}{2}(V_a^x+f(x)-f(a))$$ $$n(x)=\frac{1}{2}(V_a^x+f(x)-f(a))$$ S0 $f(x)=p(x)-n(x)+f(a)$ ,$V_a^x=p(x)+n(x)$ $$\int_a^b\vert f'(t) \vert dt=\int_a^b\vert p'(t)-n'(t) \vert dt $$ $$V_{a}^x=\int_a^b\vert p'(t)+n'(t) \vert dt$$ So if ① holds,we must have $$\int_a^b\vert p'(t)-n'(t) \vert dt=\int_a^b\vert p'(t)+n'(t) \vert dt$$ since both $p'(t),n'(t) $is larger than $0$can we prove that min{$ p'(t),n'(t)$} $=0$ a.e.?

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Maybe I got everything wrong but you've said that you can prove that $$\int_a^b\vert f'(t) \vert dt \le V_{a}^b(f)$$ If it is so, the opposite inequality is quite simple: let the $x_0=a<x_1<x_2<...<x_n=b$ is a partition of the segment $[a;b]$ Then $$\sum_{k=0}^{n-1} |f(x_{k+1})-f(x_{k})| = \sum_{k=0}^{n-1}|\int_{x_k}^{x_{k+1}} f'(t)dt| \le \sum_{k=0}^{n-1}\int_{x_k}^{x_{k+1}} |f'(t)|dt = \int_a^b |f'(t)|dt$$ And taking a supremum of both parts gives you $$ V_{a}^b(f) \le \int_a^b\vert f'(t) \vert dt $$

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  • $\begingroup$ Thank you!For almost every $x_0\in [a,b]$,$\vert f'(x_0) \vert=\lim_{h_n\to 0} \vert \frac{(x_0+h_n)-f(x_0)}{h_n} \vert \le \lim_{h_n\to 0} \frac{V_a^{x_0+h_n}(f)-V_a^{x_0}(f)}{h_n}$ then we can get the desire inequality. $\endgroup$ – J.Guo Oct 5 '18 at 12:14
  • $\begingroup$ By the way,can we prove that $\frac{d}{dx}V_a^x= \vert f'(x) \vert$? $\endgroup$ – J.Guo Oct 5 '18 at 12:17
  • $\begingroup$ Well, isn't it clear? As soon as $\int_a^b\vert f'(t) \vert dt=V_{a}^b(f)$ then for every $x \in [a;b] \Rightarrow \int_a^x\vert f'(t) \vert dt=V_{a}^x(f)$ $\endgroup$ – Anton Zagrivin Oct 5 '18 at 12:26
  • $\begingroup$ And when you do $\frac{d}{dx}$ to the left part, you get $\frac{d}{dx}V_a^x$, and when to the right, you get $\vert f'(x) \vert$ almost everywhere. Or you need to know why(about the right part?) $\endgroup$ – Anton Zagrivin Oct 5 '18 at 12:28
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    $\begingroup$ I could see it now ,thanks a lot. $\endgroup$ – J.Guo Oct 5 '18 at 12:40
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I don't quite understand your notation, you mean that $V_a^x$ is a total variation of $f$ on the segment $[a;x]$ ? If so, your statement $$\int_a^b\vert f'(t) \vert dt=V_{a}^x$$ seems strange because the left part of it is just a fixed real number and a right part is a function of $x$ (if I got it right and $V_a^x\overset{def}{=}V_a^x(f)$

What is really can be proved is that if $f$-absolutely continious on $[a;b]$, then $$\int_a^b\vert f'(t) \vert dt=V_{a}^b(f)$$

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  • $\begingroup$ Sorry,that is what I really want to know.How to prove $\int_a^b\vert f'(t) \vert dt=V_{a}^b(f)$ $\endgroup$ – J.Guo Oct 5 '18 at 11:25

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