# A nowhere continuous derivative?

August 1, 2008 7:23 PM Subscribe

Is there a differentiable function f(x) such that the derivative is nowhere continuous? That is, f'(x) has no point of continuity?

Googling has led me to believe this does not exist, though I can't find anything like a proof, only vague references to Baire classes.

Can you tell me where I would find a proof?

Googling has led me to believe this does not exist, though I can't find anything like a proof, only vague references to Baire classes.

Can you tell me where I would find a proof?

But itsIf it is differentiable at a point, then it is also continuous at said point.

*derivative*is

*not*necessarily continuous at that point - there are definitely functions that are differentiable but whose

*derivatives*are not continuous.

He's asking if there are differentiable functions whose

*derivatives*are nowhere continuous, not if there are differentiable functions that are nowhere continuous.

posted by Flunkie at 8:32 PM on August 1, 2008

Well, I got curious about such an interesting question and found this. As my knowledge and courses in calculus never went into such arcane things as Baire classes, I'll just point you that way.

posted by Iosephus at 8:49 PM on August 1, 2008

posted by Iosephus at 8:49 PM on August 1, 2008

Sorry, I just realized that you probably already saw that one in your googling and I'm being less than helpful. Before embarrassing myself any further, did you also see this Yahoo! Answers one, that seems to be a rephrasing (or particular case?) of your question. It does sound like some advanced real analysis would be involved in your proof...

posted by Iosephus at 9:28 PM on August 1, 2008

posted by Iosephus at 9:28 PM on August 1, 2008

And finally: is this proof of any help? It seems to require (again) some advanced real analysis background in which I'm lacking to properly judge.

posted by Iosephus at 9:39 PM on August 1, 2008

posted by Iosephus at 9:39 PM on August 1, 2008

The first thing that popped into my head is the Weierstrass Function. It's everywhere continuous, but nowhere differentiable, as the derivative diverges at all points. I'm not sure it's quite what you're looking for, but it may give you some ideas.

posted by PaperDragon at 9:57 PM on August 1, 2008

posted by PaperDragon at 9:57 PM on August 1, 2008

Shouldn't this be equivalent to asking whether there exists any nowhere-continuous function that is (Riemann) integrable? If so, then the answer is no since it is necessary that a function be mostly continuous in order for it to be integrable. People seem to discuss a proof of this here.

posted by sentient at 10:14 PM on August 1, 2008

posted by sentient at 10:14 PM on August 1, 2008

From the geometric interpretation of a derivative it seems intuitively that the derivative must be continuous at least from one direction where it is defined. (Of course intuition and mathematics can often collide.)

Since the derivative is the gradient of the chord connecting two points on a curve as the length of the chord approaches zero, it would seem to imply that the derivative is continuous over that region. Since a differentiable function must be continuous between x and x+h, the derivative must also be defined between x and x+h. In other words, if it is differentiable at x, it must also be differentiable at h. The gradient of the chord must smoothly or continuously approach the gradient of the tangent.

posted by JackFlash at 11:07 PM on August 1, 2008

Since the derivative is the gradient of the chord connecting two points on a curve as the length of the chord approaches zero, it would seem to imply that the derivative is continuous over that region. Since a differentiable function must be continuous between x and x+h, the derivative must also be defined between x and x+h. In other words, if it is differentiable at x, it must also be differentiable at h. The gradient of the chord must smoothly or continuously approach the gradient of the tangent.

posted by JackFlash at 11:07 PM on August 1, 2008

**Iosephus**'s last link should have it. My notes are in italic below:

For a function g: R \to R let Cont(g) denote the set of continuity points of g.

A function g:R\to R that is the _pointwise_ limit of a sequence of

continuous functions is called a Baire 1 function.

*This isn't the actual important thing we need - we need a different result that I mention below, but its nice to have the definition around*

The following result is a classical result (going back to Baire), cf. [K,

(24.14)]

*I haven't actually looked into this stuff. The citation would have the proof of this theorem, though.*

Theorem. If f:R\to R is Baire 1, then Cont(f) is dense in R.

Assume there was a continuous function f for which f' was discontinuous everywhere (i.e., Cont(f) = \emptyset). Then Cont(f) would not be dense in R, and f' is not Baire 1. There appears to be a theorem, for which I do not have the citation or the proof, that says a derivative of a differentiable function is either Baire 0 or Baire 1. Since Baire 0 is continuous, f' is clearly not Baire 0 either, so f is not continuous, so contradiction.

Assume there was a continuous function f for which f' was discontinuous everywhere (i.e., Cont(f) = \emptyset). Then Cont(f) would not be dense in R, and f' is not Baire 1. There appears to be a theorem, for which I do not have the citation or the proof, that says a derivative of a differentiable function is either Baire 0 or Baire 1. Since Baire 0 is continuous, f' is clearly not Baire 0 either, so f is not continuous, so contradiction.

posted by devilsbrigade at 11:43 PM on August 1, 2008

(replace Cont(f) with Cont(f') in the first two lines of the last paragraph). I should also point out that 'Baire 0 is continuous' is by definition, not something I'm claiming.

posted by devilsbrigade at 11:46 PM on August 1, 2008

posted by devilsbrigade at 11:46 PM on August 1, 2008

And as a final note, thank you for posting this problem! To think this night had almost gone to waste...

posted by devilsbrigade at 11:47 PM on August 1, 2008

posted by devilsbrigade at 11:47 PM on August 1, 2008

That's incorrect.From the geometric interpretation of a derivative it seems intuitively that the derivative must be continuous at least from one direction where it is defined.

That's correct.(Of course intuition and mathematics can often collide.)

The classic example is the function f where:

f(0) = 0

f(x<>0) = x^2 * sin ( 1 / x )

This function has a derivative at all points, but its derivative is discontinuous at zero:

f'(0) = 0

f'(x<>0) = 2x * sin ( 1/ x ) - cos ( 1 / x )

See here (PDF) for details.

posted by Flunkie at 12:03 AM on August 2, 2008 [1 favorite]

Theorem 9.9 (from Principles of Real Analysis, by Aliprantis and Burkinshaw):

Let $X$ be a topological space, and let ${f_n}$ be a sequence of $C(X)$. If ${f_n}$ converges point-wise to a real-valued function $f$, then the set $D$ of all points of discontinuity of $f$ is a meager set.

That should do it.

posted by metastability at 9:11 AM on August 2, 2008

Let $X$ be a topological space, and let ${f_n}$ be a sequence of $C(X)$. If ${f_n}$ converges point-wise to a real-valued function $f$, then the set $D$ of all points of discontinuity of $f$ is a meager set.

That should do it.

posted by metastability at 9:11 AM on August 2, 2008

To elaborate, if $g$ is your differentiable function, then the hypotheses of the above theorem hold for the derivative of $g$, where the sequence is given by $$f_n(x) = (g(x + delta_n) - g(x)) /delta_n$$ for some partition of the real line.

posted by metastability at 9:18 AM on August 2, 2008

posted by metastability at 9:18 AM on August 2, 2008

Thanks to

posted by exphysicist345 at 3:57 PM on August 2, 2008

**PaperDragon**for the link to the**Weierstrass Function.**The key there is that the graph of the WF looks like a fractal pattern — that is, "The function has detail at every level, so zooming in on a piece of the curve does not show it getting progressively closer and closer to a straight line. Rather between any two points no matter how close, the function will not be monotone."posted by exphysicist345 at 3:57 PM on August 2, 2008

This thread is closed to new comments.

posted by Loto at 7:57 PM on August 1, 2008