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Problems relating to Absolute Continuity

  1. Aug 15, 2007 #1
    Hi, it's been awhile since I have studied Lebesgue measure so I'm trying to re-learn the material on my own. Most of my friends don't remember much as well so it's been a bit of a struggle trying to work on these problems on my own. Thank you for any kind of help!

    OMIT Question 1. If [itex]f:[a,b]\rightarrow \mathbb{R} [/itex] is absolutely continuous and one-to-one, then is [itex]f^{-1}[/itex] absolutely continuous? If so, prove. If not, provide a counterexample.


    OMIT Question 2. If [itex]f:[a,b]\rightarrow [c,d] [/itex] is 1-1 and absolutely continuous, then
    OMIT a. is [itex]f^{-1}[/itex] of bounded variation on [itex][c,d] [/itex]?
    OMIT b. if [itex]E \subseteq [c,d] [/itex] and [itex]m(E)=0[/itex], then do you think [itex]m(f^{-1}(E))=0[/itex]? If I know that [itex]f^{-1}[/itex] is absolutely continuous, then I can prove that [itex]f^{-1}[/itex] sends sets of measure zero to sets of measure zero....




    Question 3. Let [itex] f:[0,1] \rightarrow\mathbb{R} [/itex] satisfy [itex] |f(x) - f(y)|\leq |x^{1/3}-y^{1/3}| [/itex] for all [itex]x,y \in [0,1][/itex]. Must [itex]f [/itex] be absolutely continuous? Justify.

    My question 3 is hard....



    Edit: I removed the last problem which I eventually figured out...
     
    Last edited: Aug 16, 2007
  2. jcsd
  3. Aug 15, 2007 #2
    Definition and Theorem

    By the way, here are a definition and a theorem in case one has forgotten some terms relating to absolute continuity.


    Definition: We say f is absolutely continuous if for any [itex]\epsilon > 0[/itex] there is some [itex]\delta > 0 [/itex] so that

    [itex]\sum_{j=1,...,N} \left| f(b_j)-f(a_j) \right| < \epsilon[/itex] whenever [itex]\left{ \sum_{j=1,...,N} (b_j-a_j) < \delta \right} [/itex], where the intervals [itex](a_j, b_j) [/itex] are disjoint.


    Theorem: If f is absolutely continuous on [a,b], then f' exists almost everywhere and [itex]f \in L^1([a,b]) [/itex]. Moreover, for [itex]a \leq x \leq b[/itex],

    [itex] \int_a^x f'(y) dy = f(x) -f(a)[/itex].


    A few other tools/facts: If f is absolutely continuous, then f is of bounded variation. If f is Lipschitz continuous, then f is absolutely continuous. If f is absolutely continuous, then f is uniformly continuous and continuous.


    The reason why I'm having a hard time with these problems is because I don't know enough theorems relating to absolute continuity. So I don't have enough tools which I can apply to these problems.

    I'll think about these some more...
     
  4. Aug 16, 2007 #3

    mathwonk

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    absolute continuity is the property you need to insure that having derivative = 0, a.e., makes the function constant, i believe. thats what it is realy for, to generalize the MVT, and hence make integration possible. so to me these questions seems pretty artificial and technical, if i may say so.
     
    Last edited: Aug 16, 2007
  5. Aug 16, 2007 #4
    Thanks for your reply Mathwonk! And thanks for all the other comments from the other (algebra) post. I haven't been able to keep up with all the comments because I'm currently reviewing analysis. But I will read them in next week's time.

    I figured out all of the above except for my Question 3. I can type what I came up with so far (a partial proof), but I can't finish it or can I come up with a counterexample...


    By the way, above comment is a good intuition! It helps to understand absolute continuity better!
     
    Last edited: Aug 16, 2007
  6. Aug 16, 2007 #5
    Let's modify the above problem and assume [itex] |f(x) - f(y)|\leq |x^{1/2}-y^{1/2}| [/itex] instead. I don't know if this is easier.

    Proof: Let [itex]\cup_{i=1}^N (a_i, b_i)[/itex] be a disjoint union of open intervals in [0,1] such that [itex]\sum_{i=1}^n (b_i-a_i) < \delta [/itex].

    Then [itex]\sum_{i=1}^N |f(b_i)-f(a_i)|\leq \sum_{i=1}^N |(b_i)^{1/2}-(a_i)^{1/2}| = \sum_{i=1}^N | b_i-a_i |/|(b_i)^{1/2}+(a_i)^{1/2}| [/itex]

    I want to show that this is less than epsilon.

    Since [itex] (a_i, b_i)[/itex] is an open interval, [itex]a_i \not = b_i[/itex] right? So we have [itex]\sum_{i=1}^n b_i+ a_i > c [/itex] for some c...

    I would like to find a lower bound...
     
  7. Aug 16, 2007 #6

    mathwonk

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    i guess for 3, i would take f = x^1/3 and see if it is lipschitz, probably not. then try to see if abs contin.

    this also suggests answers to questions about inverses of abs cont fcns.

    and read the lebesgue decomposition lemma.?

    it seems also that an absolutely continuous fcn is the integral of its derivative. now the derivative of x^1/3 is unbounded but has an improper integral.

    i dont know how to use this. probably a little reading or googling is in order.

    well wiki says local absolute continuity is equivalent to absolute continuity of the distribution derivative. so the unboundedness of the derive may imply x^1/3 is not abs contin. see royden if available. this is my weakest basic pure subject.
     
    Last edited: Aug 16, 2007
  8. Aug 16, 2007 #7

    quasar987

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    Here's another thm for you: Differentiable ==> Lipschitz continuous.

    So if you want to follow mathwonk's program, you just have to investigate lipschitz zness at x=0. (it probably isn't)
     
  9. Aug 16, 2007 #8

    mathwonk

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    it does not have to be diffble everywhere, just a.e.

    and i think the result is that continuous derivative implies lipschitz.

    but check out this page.

    http://planetmath.org/encyclopedia/FundamentalTheoremOfCalculusForTheLebesgueIntegral.html

    it all depnds on whether x^(-2/3) is L1 on [0,1] or not and represents x^(1/3) as an integral which i think it does, in the lebesgue sense.

    i.e. x^1/3 does seem to be the lebesgue integral of its unbounded derivative. so then one just applies the defn of abs con to get the answer yes? to prob 3?
     
    Last edited: Aug 16, 2007
  10. Aug 16, 2007 #9

    quasar987

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    Allow me. :biggrin:

    Since f is differentiable at x, given e>0, there is a d>0 such that |h|<d ==>

    [tex]|f(x+h) - f(x) - f'(x)h| \leq \epsilon |h|[/tex]

    Thus, by the "modified" triangle inequality,

    [tex]|f(x+h) - f(x)| \leq |f'(x)h|+\epsilon |h|=(|f'(x)|+\epsilon)|h|[/tex]

    This is the inequality required for Lipschitz continuity at x.
     
  11. Aug 17, 2007 #10
    Thanks for all your input!!

    I was joyful at first but I don't think f is Lipschitz continuous at x for almost all x implies that f is absolutely continuous.

    Maybe f is Lipschitz continuous at x for almost all x implies that f is almost absolutely continous?
     
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