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Is the anti-derivative of a continuous function continuous?

  1. Nov 28, 2009 #1
    Hi guys

    I have been wondering: Say we have a continuous function f. I integrate f to obtain its anti-derivative called capital f, i.e. F. Now I wish to prove the differentiability of F, and in order to do so, I need the fact that F is continuous (this is just something I need in my proof).

    Now, the problem is that I cannot deduce continuity of F on the fact that F is differentiable, since I wish to prove the differentiability of F. What can I do instead?

    I thought of using the argument (which I am not sure is correct) that the integral of a continuous function is continuous. Am I allowed to do this?
    Last edited: Nov 28, 2009
  2. jcsd
  3. Nov 28, 2009 #2


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    How would you argue for it? :smile:

    Try to make a proof for that assertion!

    By the way, yes, it is provably correct..:smile:
  4. Nov 28, 2009 #3
    Hmm, I can't come up with a proof for it, but I mean: Why shouldn't it be continuous? If we look at it as describing the area under a continuous graph, then needless to say, it has to be continuous. But again, this is hardly a proof..
  5. Nov 28, 2009 #4
    Isn't the fact that f is integratable (to F) enough to show that F is differentiable?
  6. Nov 28, 2009 #5
    I am told to do it "the long way" :smile:
  7. Nov 28, 2009 #6
    There's something important you can say about the anti-derivative. What is it? The answer is in the name.
  8. Nov 29, 2009 #7
    I cannot use that F is an anti-derivative (i.e. that it is differentiable) to prove its continuity, since I am trying to prove its differentiability. So I must use another argument.
  9. Nov 29, 2009 #8


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    If F is the antiderivative of f then F'=f, so obviously F is in C^1.
  10. Nov 29, 2009 #9
    I am quoting what I said earlier:

  11. Nov 29, 2009 #10


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    The whole point of being an "anti-derivative" is that it has a derivative! How else are you going to use the fact that this function is the anti-derivative of some function?
  12. Nov 29, 2009 #11
    But isn't that a circular argument? I mean, I am trying to prove differentiability, so I can't argue from this "ansatz"/claim that it is continuous.
    Last edited: Nov 29, 2009
  13. Nov 29, 2009 #12


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    Alright, then, how are you defining "anti-derivative"? After all, if you are going to prove anything about an anti-derivative, you are going to have to use the fact that is is an anti-derivative- you are going to have to use the definition.

    The definition I know of "anti-derivative" is "F(x) is an anti-derivative of f(x) if and only if F'(x)= f(x)". It pretty much follows from the fact that f(x) is the derivative of F(x) that F(x) is differentiable. And, of course, "differentiable" is stronger than "continuous". If F(x) is differentiable, it must be continuous.

    The anti-derivative is a "smoothing" operation. F(x) is always at least as "smooth" as f(x). Even if f(x) is not continous, F(x) is. And F(x) is "continuously differentiable" wherever f(x) is continous.
    Last edited by a moderator: Nov 29, 2009
  14. Nov 29, 2009 #13
    I would to it like you did. But you have a good point.

    Thanks to all for helping.
  15. Nov 29, 2009 #14
    since the integral over a interval [x,x] is always zero (given the integrand is continuous)
    then take the limit of the next expression:

    [tex] F(x+h)-F(x)=\int^{x+h}_{0}f(x)dx-\int^{x}_{0}f(x)dx=\int^{x+h}_{x}f(x)dx[/tex]
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