# Thread: Derivative with finite jump discontinuity?

1. ## 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.

2. Originally Posted by Mazerakham
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

3. 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 $\displaystyle x\ne a$ 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 $\displaystyle x\ne b$ 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.

4. Wow, that's great. Yep, that (just about) gets rid of the possibility of a jump discontinuity. Thanks!

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# jump discontinuity differentiation

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