Do derivatives have to be continuous anywhere?

Question

Suppose a real-valued function f is continuous on a closed interval [a, b] and differentiable on the open interval (a, b).

I know that the derivative f' does not have to be continuous at every point (for example, x^2 sin (1/x) is differentiable at 0, but its derivative is not continuous there).

Does f' have to be continuous anywhere?

If it does, what is the most "non-continuous" it can be? For example, can f' be discontinuous at uncountably many points? Can it be discontinuous on a set of positive measure? Can it be discontinuous almost everywhere?

If not, can someone point me to an example of a function which is differentiable on an open interval, but its derivative is not continuous anywhere on that interval?

Thanks for your help.

Additional Details

By the function x^2 sin (1/x) above, I of course mean that the function is "patched" so that its value is 0 when x = 0. (The "patched" function is differentiable on any open interval containing 0.)

whitesox08: I am not asking if f is continuous. Of course it is if it is differentiable. I am asking where, if anywhere, the derivative f' must be continuous.

M4tr!x, I've given an example of a function which is differentiable everywhere, but has non-continuous derivative. That function is (defined piecewise):

f(x) = x^2 sin (1/x), if x is not 0
f(x) = 0, if x is 0

This function is continuous everywhere, and differentiable everywhere, and its derivative is NOT continuous at 0. (Of course, to differentiate this function at 0, you need to use the definition of the derivative, because the usual formulas won't work at 0.)

Again, if a function f is differentiable, then of course it must be continuous. My question asks, instead, whether its *derivative* f' has to be continuous. I already know it doesn't have to be continuous everywhere, by the example above. I want to know if it can be non-continuous everywhere or not.

Answer 1

Best Answer - Chosen by Asker

The "Cantor's staircase" function has a derivative that is discontinuous on the Cantor set. See: http://www.cut-the-knot.org/do_you_know/…

A necessary and sufficient condition that a set A in some interval [a,b] is a set of continuity of a derivative (of some function on [a,b]) is that A is a dense set of type Gδ.

In other words, the complement of the set of discontinuities must be dense, and it must be Gδ.

That's really all there is to it, that's the exact answer to your question (I don't think I see it in anyone else's answer so far), so I won't go on at length about other conclusions/theorems - others seem to have covered that.

Recall: Gδ means countable intersection of open sets.

Edit: response to Steiner. What do you mean "interesting"? I just provided the necessary and sufficient conditions, which are really the only way to answer this question completely. That's a bit more than an interesting side note.

Source(s):

http://books.google.com/books?id=fXfEG-F…
http://en.wikipedia.org/wiki/G-delta_set

Asker's Rating:

Asker's Comment:

Thanks! This is exactly what I wanted to know. Steiner, your answer was very helpful and informative also, but I had to pick one. Again, thanks for all the info.

Answer 2

the derivative must be a continuous function to be valid. (continuous everywhere) Derivatives do not exist for discontinuous functions. For the x^2 sin (1/x), x at 0 is undefined for the original function. You cannot have a 0 in the denominator.
Refer to Rolles Theorem also, which validates your very first statement in the question.
Basically, you cannot differentiate at a point where the original function is discontinuous at that point. In a generality, you cannot differentiate a function at a point where it is discontinuous. So, you can differentiate at certain points where the function is NOT discontinuous.
There is no such function which is "differentiable on an open interval but its derivative is not continuous anywhere on that interval."

for example: f(x) = 1/x
in terms of x so d/dx you CAN differentiate to get -x^-2
but if you were differentiating at a certain point that happened to be x = 0, then f'(0) does not exist because f(0) is undefined and thus discontinuous at x = 0.

Answer 3

I'd say that a function f that is differentiable everywhere might have a nowhere differentiable derivative f'. Here is why:

Let g(x) be the Weierstrass zeta function. This function is continuous everywhere but differentiable nowhere.
Construct the function F(x)= integral from 1 to x of g(t)dt.
By the fundamental theorem of calculus F is differentiable everywhere and its derivative is g(x) - a nowhere differentiable function!

Hope this makes sense. And oh, by the way, here is some information about this zeta function:

http://en.wikipedia.org/wiki/Weierstrass…

Answer 4

As your example shows( provided you define f(0) = 0), f' does not have to be continuous everywhere. But, as for your questions, let D be the set of discontinuities of f' on an interval I:

1) D is meager in Baire classification, that is, it is contained in a countable union of nowhere dense sets. This is consequence of the fact that f' is the limit of a sequence of continuous functions defined on I. The discontinuities of limits of continuous functions form a meager set.

2) (1) implies that D has an empty interior. Therefore, the set of continuities of f' is dense in I.

3) D can have positive measure. It can even have full measure, that is, measure(D) = lenght(I) (possibly oo if I is all of R). So, D can very well be uncountable. Its complement D', the set of continuities on I, is ALWAYS uncountable.

4) The conclusions cited above show there's NO function differentiable on an open interval I but such that f' is nowhere continuous on I. So, it's impossible to give the example you mention at the end of your question.

5) Even if not continuous everywhere on I, derivatives always present the intermediate value property: If f is differentiable on [a, b], then f' assumes on (a,b) every value between f'(a) and f'(b). This is known as Darboux Theorem, and the usual proof is done considering the function g(x) = f(x) - k*x, where k is a number between f'(a) and f'(b).

6) An immediate consequence of (5) is that, if f' is monotone on I, then f' is continuous on I

7) Derivatives NEVER present jump discontinuities. In addition, if f' is discontinuous at a, then f' does NOT have a limit at a.

8) There's an interesting theorem related to derivatives: If f is continuous at a+, f' exists on (a, a + d) for some d >0 and lim x --> a+ f'(x) = L, then f is differentiable at a+ and f'(a+) = L, which implies f' is continuous at a+. Of course, similar conclusions hold for a- and a, if the given conditions hold to the left and to the right of a. (this theorem implies (7))

Well, that's what I remenber right now. Hope this helps a bit. Sorry if I went a bit off topic.

EDIT:

Another interesting conclusion, not directly related to your questions but maybe worth mentioning: If f is differentiable on a interval I, then there exists a subinterval of I where f is Lipschitz. Actually, it's not necessary that f be fully differentiable on I, it suffices all of it's Dini's derivatives exist on I. This is NOT the same as to say f is locally Lipschtiz on I, is not so strong.

EDIT2

As for cheeser remark, I think it's interesting to point out that, if f is any function defined in any topological space and with values in a metric space, then the set of points at which f is continuous is a Gδ, which implies the set of its discontinuities is an Fσ. If f is the derivative of some function, then, in addition to being Fσ, the set of discontinuities has an empty interior, that is, it's meager.

Source(s):

In the group sci.math, of Google, there's an intersting post on this subject, with several sources. If you're interested, I'll look it up and send the link to you.