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Inequality integral absolute value derivative

  1. Nov 17, 2013 #1
    Here's a claim: Assume that a function [itex]f:[a,b]\to\mathbb{R}[/itex] is differentiable at all points in its domain. Then the inequality

    |f(b) - f(a)| \leq \int\limits_{[a,b]}|f'(x)|dm(x)

    holds. The integral is the Lebesgue integral.

    Looks simple, but I don't know if this is true. There exists functions, which are differentiable everywhere, but not monotonous on any interval. The result is related to Baire category theorem. Therefore we cannot assume that there would be intervals where the derivative has a non-changing sign, and I don't know where to start the proving.
  2. jcsd
  3. Nov 17, 2013 #2
    f and |f| are continuous functions on a compact interval, hence are Riemann integrable. The Lebesgue integral is the Riemann integral in this case. By the fundamental theorem of calculus, [tex]|f(b) - f(a)| = |\int\limits_{[a,b]}f'(x)dm(x)| \leq \int\limits_{[a,b]}|f'(x)|dm(x)[/tex]
  4. Nov 17, 2013 #3
    The assumptions of the fundamental theorem of calculus are not available here, since the derivative is not necessarily Riemann integrable.
  5. Nov 17, 2013 #4
  6. Nov 17, 2013 #5
    Oops. My bad!
  7. Nov 18, 2013 #6
    Henstock-Kurzweil integral is always motivated by examples like this, and I have never seen to point. Truth is that you don't need more advanced integral to integrate this, because it can be done with the ordinary Riemann or Lebesgue integrals by using the limit [itex][\delta,1]\to [0,1][/itex] in the domain of integration.

    This looks interesting. If the Henstock-Kurzweil integral can be used to prove my claim, it would be the first time I see something useful coming out of it. Unfortunately it is still not clear to me how this happens, since the Wikipedia page doesn't contain proofs.
  8. Nov 18, 2013 #7
    If [itex]a=x_0<x_1<\cdots<x_n=b[/itex], notice that [itex]|f(a)-f(b)|\leq\sum_{i=1}^n |f(x_i)-f(x_{i-1})|[/itex]. This may be of some use to you.
  9. Nov 18, 2013 #8
    The integral


    is the supremum of all integrals of simple non-negative functions below the mapping [itex]x\mapsto |f'(x)|[/itex]. The sum

    \sum_{i=1}^n (x_i-x_{i-1})|f'(\xi_i)|

    where [itex]x_{i-1}\leq\xi_i\leq x_i[/itex] represents an integral of a simple function that is not necessarily below the mapping [itex]x\mapsto |f'(x)|[/itex], so I don't see how the previous remark could be of some use here.

    Don't underestimate the original problem. Just post your proof, if you have it. This is not a homework style question.
  10. Nov 18, 2013 #9
    Let [itex]K=|f(b)-f(a)|[/itex]

    Given a finite partition [itex]\pi=\{a=x_0^\pi<x_1^\pi<\cdots<x_{n^\pi}^\pi=b\}[/itex] of the interval [itex][a,b][/itex], there exists (by mean value theorem) for each [itex]i\in\{1,...,n^\pi\}[/itex] some [itex]z_i^\pi\in[x_{i-1}^\pi,x_i^\pi][/itex] with [itex](x_i^\pi-x_{i-1}^\pi)f'(z_i^\pi) = f(x_i^\pi)-f(x_{i-1}^\pi)[/itex]. Then we can bound the Riemann sum [tex]S^\pi := \sum_{i=1}^{n^\pi}(x_i^\pi-x_{i-1}^\pi)|f'(z_i^\pi)|[/tex] (for [itex]|f'|[/itex] from [itex]a[/itex] to [itex]b[/itex]) by observing that [tex]S^\pi = \sum_{i=1}^{n^\pi}\bigg|(x_i^\pi-x_{i-1}^\pi)f'(z_i^\pi)\bigg|= \sum_{i=1}^{n^\pi}\bigg|f(x_i^\pi)-f(x_{i-1}^\pi)\bigg| \geq \bigg|f(x_{n^\pi}^\pi)-f(x_0^\pi)\bigg|=K.[/tex] So let's assess our progress. We've produced, for any finite partition of the interval, a Riemman sum with respect to that partition which is [itex]\geq K[/itex].

    This is a full proof in the case of [itex]|f'|[/itex] Riemann integrable, but it's still a useful step in the general case of of [itex]|f'|[/itex] measurable. To extend, Lusin's theorem might be helpful.
    Last edited: Nov 18, 2013
  11. Nov 19, 2013 #10
    If it is possible to prove the claim with Henstock-Kurzweil integral, it is probably also possible to produce a more primitive proof by using the similar ideas that are present in the Henstock-Kurzweil integral. I would prefer such way.

    Let [itex]\epsilon > 0[/itex] be fixed. Is it possible to find a simple function [itex]y:[a,b]\to\mathbb{R}[/itex] such that

    y(x) = \sum_{i=1}^N y_i \chi_{\Omega_i}(x)

    (where [itex]y_i[/itex] are real numbers, [itex]\chi[/itex] is the indicator function, and [itex]\Omega_1,\ldots,\Omega_N[/itex] are measurable subsets of [itex][a,b][/itex].), [itex]0\leq y(x) \leq |f'(x)|[/itex] and

    |f(b) - f(a)| < \sum_{i=1}^N y_i m(\Omega_i) + \epsilon.

    If this [itex]y[/itex] can be found, the desired result follows by the definition of the Lebesgue integral. Clearly, we must find [itex]y[/itex] that obtains somehow large values for this inequality to come true. One possible way to start is to define the following sets

    \Psi_{+,n,m} = \Big\{ x\in[a,b]\;\Big|\; \frac{n}{m}\leq f'(x)< \frac{n+1}{m}\Big\}
    \Psi_{-,n,m} = \Big\{ x\in [a,b]\;\Big|\; -\frac{n+1}{m}< f'(x) \leq -\frac{n}{m}\Big\}

    for positive integers [itex]n,m[/itex]. These sets and numbers [itex]\frac{n}{m}[/itex] can be used to define all kinds of simple functions that come close to [itex]|f'(x)|[/itex] from below.

    But I don't see how to continue from here. The sets [itex]\Psi_{\pm,n,m}[/itex] do not necessarily contain any intervals, and I cannot use the mean value theorem here. Also, the sets do not produce anything that would allow the use of triangle inequality for [itex]|f(b)-f(a)|<\cdots[/itex]. So this is where stuff gets stuck.
  12. Nov 19, 2013 #11


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    If the function is oscillating back and forth so fast that it's not even increasing or decreasing in any interval, it seems like the problem statement should be more obviously true, not less obviously (heuristically only of course). The value of f(x) will change little as x sweeps across the interval, but we will be accumulating a large amount of |f'(x)|.

    From this I would assume the best approach is to approximate f'(x) with some nicer functions and take a limit, but it's hard to see how exactly this can be done without accidentally trying to prove the fundamental theorem of calculus for arbitrary differentiable functions under the Lebesgue integral.
  13. Nov 19, 2013 #12
    Yes, I'm aware of this paradox in the difficulty. It is probably quite common with strange functions that the integral of the absolute value of the derivative is actually infinite. But the finite possibility remains too, and for those cases, the proof remains a mystery.

    I wouldn't mind if some generalization of fundamental theorem of calculus gets proven. I know that one possible generalization to the fundamental theorem of calculus can be proven by using the claim of this thread as a lemma. That's why I'm interested in this claim, in the end.
  14. Jan 11, 2014 #13
    I have found that the proof for the claim can be found in the book A First Course in Sobolev Spaces by Giovanni Leoni. In Chapter 3.
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