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Math Help - Derivative with finite jump discontinuity?

  1. #1
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    Derivative with finite jump discontinuity?

    I haven't seen this question posed by a book. A proof or counter-example would be great for this statement:

    Let f be a function whose derivative f ' exists everywhere on a closed interval [a,b]. Then (so I claim) f ' cannot have a JUMP discontinuity.

    I believe this is true, and I included the restriction on the type of discontinuity because there would be a counterexample to the claim that " f ' must be continuous ": consider
    f(x) = [ x^2 sin(1/x) ] + x

    So my question is simply, if the derivative exists everywhere, can it have infinite or finite jump discontinuities? To define that rigorously, I am asking: is it possible that there can be a point c such that the right-handed and left-handed limits of the derivative at c exist but are not equal?

    You may note that the absolute value function fails to be a counterexample because the derivative doesn't exist at 0.
    Last edited by Mazerakham; November 27th 2010 at 08:52 PM. Reason: wanted to make the question more clear
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Mazerakham View Post
    I haven't seen this question posed by a book. A proof or counter-example would be great for this statement:

    Let f be a function whose derivative f ' exists everywhere on a closed interval [a,b]. Then (so I claim) f ' cannot have a JUMP discontinuity.
    A jump discontinuity in the derivative implies a corner for the function itself, and a function with a corner is not differentiable at the corner.

    (informally a function differentiable at a point u looks locally like a straight line at u, and so cannot have a corner)

    CB
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  3. #3
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    You can prove that if f is differentiable on interval [a, b], while f' is not necessarily continuous it must satisfy the intermediate value property on [a, b]. That is, f'(x), for x between a and b, takes on all values between f'(a) and f'(b).

    To do that, Define g(x) by g(x)= (f(x)- f(a))/(x- a) as long as x\ne a and g(a)= f'(a). Prove that g(x) is continous on [a, b] and so takes on all values between g(a)= f'(a) and g(b)= (f(b)- f(a))/(b- a). Define h(x) by h(x)= (f(b)- f(x))/(b- x) as long as x\ne b and h(b)= f'(b). Show that h(x) is continuous on [a, b] and so takes on all values between h(a)= (f(b)- f(a))/(b- a) and h(b)= f'(b). Finally, use the mean value theorem to show that each of those is a value of f'(x) for x between a and b.

    A function that has the intermediate value property cannot have a jump discontinuity.
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  4. #4
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    Wow, that's great. Yep, that (just about) gets rid of the possibility of a jump discontinuity. Thanks!
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