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.

I know there is a theorem saying if $f$ defined on $[a,b]$ is of bounded variation, then it is differentiable on $(a,b)$ a.e and $f'$ is integrable over $[a,b]$.

I wonder whether the converse is true, say, if $f$ defined on $[a,b]$ is differentiable on $(a,b)$ a.e and $f'$ is integrable, must $f$ have bounded variation?

In fact, I run into such concrete question: $$ f=x^\alpha \text{sin}\left(\frac{1}{x^\beta}\right) \text{ for $x\in(0,1]$} $$ and $f(0)=0$. The hint says that we can prove that when $\alpha > \beta$, $f$ is of bounded variation by showing $f'$ is integrable.

So here comes my question above, is such argument true?

share|cite|improve this question
@did =.=....mistake..sorry. –  hxhxhx88 May 30 '12 at 14:19

2 Answers 2

up vote 2 down vote accepted

Rudin, "Real & Complex Analysis", Theorem 7.21: If $f \in L^1[a,b]$ is differentiable at every point, then $f(x) - f(a) = \int_a^x f'(t) dt$.

Now consider $\psi_+(x) = \int_a^x \max(f'(t),0)\; dt$, $\psi_-(x) = \int_a^x \min(f'(t),0)\; dt$. Both are monotonic, hence of bounded variation. It follows that $f(x) = f(a)+\psi_+(x)+\psi_-(x)$ has bounded variation.

Note: To apply Rudin's theorem, $f$ must be differentiable everywhere, not just a.e.

share|cite|improve this answer
Thank you. BTW, do you know where can I find the proof of the formula in DonAntonio's answer? –  hxhxhx88 May 30 '12 at 15:59

We know, under the conditions in your problem, that the total variation of $f$ is given by $$V_a^b(f)=\int_a^b|f'(x)|dx$$ and since you're given $f'$ integrable then yes: $f$ is of b.v.

Added to the OP's request:

1) Under the condition of your question, in any subinterval $\,[\alpha,\beta]\subset [a,b]\,$, we have that for some $\,c\in (\alpha,\beta)\,,\,f(\beta)-f(\alpha)=f'(c)(\beta-\alpha)$

2) If $\,\mathcal P:=\{a=x_1<x_2<...<x_{n_p}=b\}\,$ is a partition of $\,[a,b]\,$, then the total variation of $f$ on this interval is defined to be $$V_a^b(f):=\sup_{\mathcal P}\sum_{k=1}^{n_p-1}\left|f(x_{i+1}-f(x_i)\right|$$

3) Can you see now from where the integral of $f'$ comes?

share|cite|improve this answer
I'm still confused. Why this formula holds? I found it in Wiki, can you give me a proof? –  hxhxhx88 May 30 '12 at 14:21
Erased and added to my answer above –  DonAntonio May 30 '12 at 16:23
Sorry, but for some extremely annoying reason every time I click on "edit" on my answer it comes up a completely messed up thing and I can't add my explanation to the OP's requeste below...I'll try later. –  DonAntonio May 30 '12 at 16:28
I suspect you need stronger conditions on $f'$ to get the integral formula for variation. Something like the set of discontinuities of $f'$ having measure $0$, so that $f'$ is Riemann integrable. –  copper.hat May 30 '12 at 18:31
Well, that's what the OP wrote: $\,f'\,$ is integrable . In fact, for the function to be of bounded variation, it's enough to require the derivative to be bounded on the interval, though of course we cannot use the integral function based only on this. –  DonAntonio May 30 '12 at 21:05

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.