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Bounded variation

  1. Jul 11, 2011 #1
    Am I right in thinking that the statement "f:[a,b]-->R is of bounded variation" is equivalent to the statements "f:[a,b]-->R has bounded range" and "f;[a,b]-->R is a bounded function".
     
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  3. Jul 11, 2011 #2

    micromass

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    Hi Rasalhague! :smile:

    No, you are not correct in thinking that. Being of bounded variation is much more strict than being bounded. For example, the function on [0,1]

    [tex]f(x)=\sin(1/x)[/tex]

    is bounded between -1 and 1, but it is not of bounded varietion. Another example is the Dirichlet function on [0,1]:

    [tex]f(x)=\left\{\begin{array}{c} 1~\text{if}~x\in \mathbb{Q}\\ 0~\text{if}~x\notin \mathbb{Q}\\ \end{array}\right.[/tex]

    This is bounded, but not of bounded variation. It can be shown that the following must hold for functions of bounded variation:

    • The function is continuous everywhere except possibly in a countable set.
    • The function has one-sided limits everywhere.
    • The function has a derivative almost everywhere (i.e. except in a set of measure 0).

    so you see that being of bounded variation is pretty strict.
     
  4. Jul 11, 2011 #3
    Argh, thanks micromass, I see now where my confusion came from. For some reason my mind was just blanking out the summation sign!! Oopsh. Ack. Blushy-faced emoticon. I think it's asking for a rest :)
     
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