Is the anti-derivative of a continuous function continuous?

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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?
 
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  • #2
arildno
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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?
How would you argue for it? :smile:

Try to make a proof for that assertion!


By the way, yes, it is provably correct..:smile:
 
  • #3
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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..
 
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Isn't the fact that f is integratable (to F) enough to show that F is differentiable?
 
  • #5
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I am told to do it "the long way" :smile:
 
  • #6
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There's something important you can say about the anti-derivative. What is it? The answer is in the name.
 
  • #7
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There's something important you can say about the anti-derivative. What is it? The answer is in the name.
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.
 
  • #8
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If F is the antiderivative of f then F'=f, so obviously F is in C^1.
 
  • #9
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I am quoting what I said earlier:


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.
 
  • #10
HallsofIvy
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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?
 
  • #11
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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?
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.
 
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  • #12
HallsofIvy
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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.
 
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Alright, then, how are you defining "anti-derivative"?
I would to it like you did. But you have a good point.

Thanks to all for helping.
 
  • #14
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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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