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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?

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  • $\begingroup$ @did =.=....mistake..sorry. $\endgroup$
    – hxhxhx88
    May 30, 2012 at 14:19

2 Answers 2

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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?

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  • $\begingroup$ I'm still confused. Why this formula holds? I found it in Wiki, can you give me a proof? $\endgroup$
    – hxhxhx88
    May 30, 2012 at 14:21
  • $\begingroup$ Erased and added to my answer above $\endgroup$
    – DonAntonio
    May 30, 2012 at 16:23
  • $\begingroup$ 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. $\endgroup$
    – DonAntonio
    May 30, 2012 at 16:28
  • $\begingroup$ 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. $\endgroup$
    – copper.hat
    May 30, 2012 at 18:31
  • $\begingroup$ 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. $\endgroup$
    – DonAntonio
    May 30, 2012 at 21:05
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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.

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  • $\begingroup$ Thank you. BTW, do you know where can I find the proof of the formula in DonAntonio's answer? $\endgroup$
    – hxhxhx88
    May 30, 2012 at 15:59

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