# f is a absolutely continuous function on $[a,b]$ prove that $①\int_a^b\vert f'(t) \vert dt=V_{a}^b$

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.?

## 2 Answers

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 $$x_0=a is a partition of the segment $$[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$$

• 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. – J.Guo Oct 5 '18 at 12:14
• By the way,can we prove that $\frac{d}{dx}V_a^x= \vert f'(x) \vert$? – J.Guo Oct 5 '18 at 12:17
• Well, isn't it clear? As soon as $\int_a^b\vert f'(t) \vert dt=V_{a}^b(f)$ then for every $x \in [a;b] \Rightarrow \int_a^x\vert f'(t) \vert dt=V_{a}^x(f)$ – Anton Zagrivin Oct 5 '18 at 12:26
• 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?) – Anton Zagrivin Oct 5 '18 at 12:28
• I could see it now ,thanks a lot. – J.Guo Oct 5 '18 at 12:40

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 $$\int_a^b\vert f'(t) \vert dt=V_{a}^b(f)$$

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