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Math people:

This question is related to how can you "fix" one of the definitions of a BV function of one variable? . Suppose $f \in BV([0,1])$. I really have two-three questions. The second only can be answered if the answer to the first is "yes". I will accept an answer to just the first if no one can answer all of them.

(1) Does $$\lim_{t \to 0^+} \frac{\int_x^{x+t} f(s)\,ds}{t}$$ exist for all $x \in [0, 1)$?

(2) If the answer to (1) is "yes", and I define $g:[0,1] \to \mathbb{R}$ by

$$ g(x) = \lim_{t \to 0^+} \frac{\int_x^{x+t} f(s)\,ds}{t} \hbox{ for } x < 1,\ g(1)= \lim_{t \to 0^+} \frac{\int_{1-t}^{1} f(s)\,ds}{t},$$

is $g=f$ Lebesgue-a.e.? And is $g$ right-continuous? A reference would be appreciated. I would guess that all are true and known, but I can't find a reference.

Stefan (STack Exchange FAN)

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Stefan Smith
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  • have you tried to apply L'hopital rule? –  Apr 12 '13 at 23:37
  • It seems to work to me, at least for (1) –  Apr 12 '13 at 23:38
  • @user67133 to make L'Hopital's Rule work, you have to show that $\lim_{t\to 0^+}\frac{d}{dt}\int_x^{x+t}f(s),ds$ exists, which is not true for general $L^1$ functions. I would guess it is true for $BV$ functions, but I am looking for a proof or a reference, since it probably has been done before. – Stefan Smith Apr 13 '13 at 00:24
  • I suppose that you have already given a look at http://www.encyclopediaofmath.org/index.php/Function_of_bounded_variation#Differentiability

    At the end there are also a lot of references for bv-functions (16 of 31 are from italians :O ). Sorry, if I can't help more :(

     –  Apr 13 '13 at 09:19
  • @user67133 : Yes, I have. I am trying to reconcile the two definitions of BV-functions of one variable. One definition is sensitive to changes of $f$ on sets of measure zero. Giusti's book says that definition can be fixed if you only consider points $x_i$ of approximate continuity of $f$. A respondent to my earlier question suggested "normalizing" $f$ so it is right-continuous - I would like to use his suggestion, but s/he did not supply enough details. – Stefan Smith Apr 13 '13 at 15:19

1 Answers1

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The answer to both questions is "yes", if you accept that $f$ is a.e. the sum of a nondecreasing function $h_1$ and a nonincreasing function $h_2$, as is well-known. Then, for all $x \in [0,1)$,

$$\lim_{t \to 0^+} \frac{\int_x^{x+t}f(s)\,ds}{t} = \lim_{t \to 0^+} \frac{\int_x^{x+t}h_1(s)\,ds}{t} + \lim_{t \to 0^+} \frac{\int_x^{x+t}h_2(s)\,ds}{t} = \lim_{x' \to x^+} h_1(x') + \lim_{x' \to x^+} h_2(x')$$

and the limits on the right exist because $h_1$ and $h_2$ are monotone. Since $h_1$ and $h_2$ are monotone, they have at most countably many discontinuities, and (2) follows.

Stefan Smith
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