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.
First, we must note that if a function is of bounded variation on $[a,b]$, then it is bounded variation on $[a,x]$ for all $a \leq x \leq b$. Furthermore, $P_a^x(f)-N_a^x(f)= f(x)-f(a)$, where $P_a^x(f)$ and $N_a^x(f)$ are the positive and negative variations of the function from a to x respectively.
Rearranging the terms we note that $f(x)= [P_x^a(f)+f(a)]- [N_x^a(f)]$ for all $x \in [a,b]$ is the difference of two increasing functions, since we're taking the variation over bigger and bigger intervals as x increases.
We can take the derivative of both sides and get $f'(x)= (P_a^x(f))'-(N_a^x(f))'$ for almost all $x \in [a,b]$, since monotone functions are differentiable almost everywhere. So, $|f'(x)| = |(P_a^x(f))'-(N_a^x(f))'|\leq (P_a^x(f))'+ (N_a^x(f))'=(T_a^x(f))'$, where $T_a^x(f)$ is the total variation of the function from a to x. Overall we have $|f'(x)| \leq (T_a^x(f))'$ almost everywhere.
Now, integrating both sides we get $\int_a^b |f'(x)| dx \leq \int_a^b (T_a^x(f))'dx$. Now since $T_a^x(f)$ is monotone increasing, we get $\int_a^b (T_a^x(f))' dx \leq T_a^b(f)-T_a^a(f) =T_a^b(f)$. This gives us the desired result $\int_a^b|f'(x)|dx\leq T_a^b(f)$.
Side note: We can show for an absolutely continuous function that the two are actually equal.