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Let $f\colon [a,b] \to \mathbb R$ be of bounded variation. Must it be the case that $|\int_a ^b f' (x) |\leq |TV(f)|$, where $TV(f)$ is the total variation of $f$ over $[a,b]$? If so, how can one prove this?

In the standard proof of the monotone differentiation theorem, it is shown tat this holds for increasing functions: if $f$ is increasing, then $\int_a ^b f'(x) \leq f(b) - f(a) = TV(f)$. I am trying to generalize this to functions of bounded variation.

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Just use the fact that a finite variation function $f$ can be decomposed as the difference of two nonnegative increasing functions, whose sum is bounded by the variation of $f$. –  George Lowther Sep 29 '11 at 0:19
It's easy to show that if $f = f_1 - f_2 $ for increasing functions $f_1$ and $f_2$, then $TV(f) \leq TV(f_1) + TV(f_2)$. However, this places an UPPER bound on $TV(f)$. I am trying to show that $TV(f)$ is greater than another quantity, so this doesn't really help here. Are you claiming that in fact $TV(f) \geq TV(f_1) + TV(f_2)$ (and therefore $TV(f) = TV(f_1) + TV(f_2)$? If so, how would I show this? –  user15464 Sep 29 '11 at 1:08
Yes, you have $TV(f)=TV(f_1)+TV(f_2)$ in the case where $f_1,f_2$ are the minimal nonnegative increasing functions with $f-f(a)=f_1-f_2$. –  George Lowther Sep 29 '11 at 1:11

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