Comments on: A nowhere continuous derivative?

Question: A nowhere continuous derivative?

thrako — Fri, 01 Aug 2008 19:23:47 -0800

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?

By: Loto

Loto — Fri, 01 Aug 2008 19:57:20 -0800

If it is differentiable at a point, then it is also continuous at said point. If the point is a discontinuity of the function f(x), then it is not differentiable at that point.

By: Flunkie

Flunkie — Fri, 01 Aug 2008 20:32:44 -0800

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

But its 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.

By: Iosephus

Iosephus — Fri, 01 Aug 2008 20:49:59 -0800

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.

By: Iosephus

Iosephus — Fri, 01 Aug 2008 21:28:16 -0800

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...

By: Iosephus

Iosephus — Fri, 01 Aug 2008 21:39:24 -0800

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.

By: PaperDragon

PaperDragon — Fri, 01 Aug 2008 21:57:43 -0800

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.

By: sentient

sentient — Fri, 01 Aug 2008 22:14:25 -0800

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.

By: JackFlash

JackFlash — Fri, 01 Aug 2008 23:07:49 -0800

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.

By: devilsbrigade

devilsbrigade — Fri, 01 Aug 2008 23:43:47 -0800

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.

By: devilsbrigade

devilsbrigade — Fri, 01 Aug 2008 23:46:53 -0800

(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.

By: devilsbrigade

devilsbrigade — Fri, 01 Aug 2008 23:47:43 -0800

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

By: Flunkie

Flunkie — Sat, 02 Aug 2008 00:03:00 -0800

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 incorrect.

(Of course intuition and mathematics can often collide.)

That's correct.

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.

By: JackFlash

JackFlash — Sat, 02 Aug 2008 01:15:20 -0800

Nice counter example, Flunkie.

By: metastability

metastability — Sat, 02 Aug 2008 09:11:40 -0800

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.

By: metastability

metastability — Sat, 02 Aug 2008 09:18:35 -0800

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.

By: exphysicist345

exphysicist345 — Sat, 02 Aug 2008 15:57:52 -0800

Thanks to 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."