Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question
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

1 Answer 1

The answer to the question posed here as to whether, for a function $f$ of bounded variation on an interval, $$\left|\int_a^b f'(x)\,dx\right| \leq \int_a^b |f'(x)|\,dx \leq V(f,[a,b]) $$ is of course yes and can be found in numerous textbooks. I don't think I need to list them. Rather more interesting is the further generalization. If $f$ is not absolutely continuous then you would have strict inequality with the variation. But what explains the difference in the values?

This was nicely done years ago by De La Vallee Poussin. His formula looks like this $$V_f(E) = V_f(E_\infty) + \int_E |f'(x)|\,dx$$ where $V_f$ is a measure that describes the total variation of $f$ on a set $E$ and $E_\infty$ are the points in $E$ at which $f'(x)=\pm\infty$. See Saks Theory of the Integral Chapter 4, Section 9 for a classical write-up of these ideas.

In terms of the title of this question (Integral of the derivative of a function of bounded variation) I thought this might be worth mentioning as otherwise someone randomly reading topics would find little else of interest here.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.