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Does a discontinuous function for have an antiderivative?

  1. Aug 7, 2013 #1
    Can F'(x) =f(x) even if f is not continuous
    I tried making a function let
    f(x) =5 if x<=5
    f(x)=4 if x>5
    f is not continuous at 5
    Then F(x) =5x x<=5
    F(x) =4x+5 x>5
    Clearly F is continuous at 5 but F is not differentiable at 5..
    So is there a discontinuous function that has F'(x) =f(x) for every x?
     
  2. jcsd
  3. Aug 7, 2013 #2

    micromass

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    Sure, some discontinuous functions have anti-derivatives. For example, take the function ##f:\mathbb{R}\rightarrow \mathbb{R}## given by

    [tex]f(x) = \sin(1/x) + \frac{\cos(1/x)}{x}[/tex]

    with ##x\neq 0## and ##f(0) = 0##. This is discontinuous, but has anti-derivative ##F(x) = x\sin(1/x)##.

    However, not all discontinuous functions admit anti-derivatives. For example, the function ##f:\mathbb{R}\rightarrow \mathbb{R}## given by ##f(x) = 0## if ##x\leq 0## and ##f(x) = 1## if ##x>0##. This can be shown to admit no anti-derivative: http://en.wikipedia.org/wiki/Darboux's_theorem_(analysis)

    The question to characterize the functions which do admit anti-derivatives seems very hard, and is (as far as I know) unsolved.
     
  4. Aug 8, 2013 #3
    What's an example of a Darboux function without an antiderivative?
     
  5. Aug 8, 2013 #4

    micromass

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    It's not difficult to show that one must exist as follows: Any function can be written as the sum of two Darboux functions. Thus if all Darboux functions had an antiderivative, then all functions have one. This is a contradiction.

    A specific example would probably be given by http://en.wikipedia.org/wiki/Conway_base_13_function This function is a Darboux function, but it is everywhere discontinuous. But it can be shown using the Baire category theorem, that any derivative is in fact continuous somewhere (and in fact, in a dense and ##G_\delta## set of points).

    This might be of interest: http://math.stackexchange.com/questions/112067/how-discontinuous-can-a-derivative-be
     
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