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Math Help - What is the derivative of total variation?

  1. #1
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    What is the derivative of total variation?

    Let f(x) be a function of bounded variation on closed, bounded interval [a, b].

    Let g(x) = V(f, [a, x])

    where V(f, [a, x]) = \sum_{k=1}^n |f(x_{i}) - f(x_{i-1}) | given some partition P = \left \{a, x_{1}, x_{2}, ...., x_{n-1}, x  \right \} of [a, x].


    How can I calculate the derivate of g(x)?
    I know of differentiation term-by-term but not sure if this would apply here.

    Thanks
    Last edited by director; March 19th 2014 at 08:54 AM.
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  2. #2
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    Re: What is the derivative of total variation?

    I would start with the definition of the derivative and see what you get:

    $\displaystyle g'(x) = \lim_{h \to 0} \dfrac{g(x+h) - g(x)}{h}$
    Last edited by SlipEternal; March 19th 2014 at 08:58 AM.
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  3. #3
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    Re: What is the derivative of total variation?

    This is what I get

    g'(x) = \lim_{h \rightarrow 0} \frac{V(f_{[a, x+h]}) - V(f_{[a, x]})}{h} = \lim_{h \rightarrow 0} \frac{V(f_{[x, x+h]})}{h} = \lim_{h \rightarrow 0} \frac{\sum_{i=1}^m|f(x_{i}) - f(x_{i-1})|}{h}

    If I consider the supremum of g' over all partitions of [a, b], is it true that |g'| \leq f'?

    For example, if my partition is just two points  P = \big\{x, x+h\big\} then I get

    g'(x) = \lim_{h \rightarrow 0} \frac{|f(x+h) - f(x)|}{h} = |f'(x)|
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  4. #4
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    Re: What is the derivative of total variation?

    No, quite the contrary. Instead, you just showed that if $g'(x)$ exists, then $g'(x) \ge |f'(x)|$ since you want the supremum over all possible partitions (one of which is the partition with just two points).
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  5. #5
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    Re: What is the derivative of total variation?

    Yes, sorry, that's what I meant to write.
    Thanks Slip.
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