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Prove that a function is not of bounded variation

  1. Dec 1, 2013 #1
    The problem statement, all variables and given/known data.

    Prove or disprove that the function ##f(x)= x^2sin^2(\dfrac{\pi}{x})## if ##0<x\leq 1## and ##f(x)=0## if ##x=0## is of bounded variation.


    The attempt at a solution.

    I've seen the graph of this function on wolfram and for me it's clearly not of bounded variation since it has too many jumps (when ##x## approaches ##0##, ##f(x)## changes its value "too quickly"). The problem is I don't know how to prove it formally.

    I've tried to prove it by contradiction: Suppose ##f(x)## is of bounded variation, then there exists ##M>0 : \sum_{k=1}^n|Δf_k|\leq M## for every partition ##π=\{x_0,...,x_n\}## of ##[0,1]##.

    Which is the proper partition ##π=\{x_0,...,x_n\}## of the interval ##[0,1]## such that ##\sum_{k=1}^n|Δf_k|> M##? I need help to find the appropiate partition that would lead me to a contradiction.
     
    Last edited: Dec 1, 2013
  2. jcsd
  3. Dec 1, 2013 #2

    Dick

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    I wouldn't be so confident that just because there are 'too many jumps' that the function isn't of bounded variation. Try estimating the position and size of all of the maxima and from that estimate the total variation.
     
  4. Dec 1, 2013 #3
    I've seen that ##f'(x)## is bounded, and this is sufficient for the function ##f## to be of bounded variation. Is this correct?
     
  5. Dec 1, 2013 #4

    Dick

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    Yes. If you'd followed my suggestion you would have seen that the total variation is approximately 2(2/3)^2+2(2/5)^2+2(2/7)^2+... which is a convergent series. So even though there are an infinite number of bumps, they still sum to a finite number. That's maybe hard to turn into a rigorous proof because the those maxes are only approximate. But they are a very good approximation. So, yes, the easiest way to prove it is to note f'(x) exists and it's bounded.
     
  6. Dec 1, 2013 #5
    Yes, sorry for not following your suggestion but I found it easier to prove that the derivative is bounded. Thanks!
     
  7. Dec 1, 2013 #6

    Dick

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    No need for apologies. I didn't mean 'prove it that way'. I just meant it to show how you can have an infinite number of jumps and still have bounded variation.
     
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