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        riddles >> putnam exam (pure math) >> Continuity & the derivative
        
(Message started by: Icarus on Mar 31st, 2004, 6:49pm)
Title: Continuity & the derivative
Post by Icarus on Mar 31st, 2004, 6:49pm
Suppose a real-valued function f is differentiable on some neighborhood of a [in] [bbr]. Must df/dx be continuous at a? If not, what restrictions can be placed on the discontinuity?
Title: Re: Continuity & the derivative
Post by Pietro K.C. on Apr 2nd, 2004, 12:16pm
Is f differentiable at a itself? For example, is the function

f(x) = | x |

being considered here?
Title: Re: Continuity & the derivative
Post by Icarus on Apr 3rd, 2004, 7:08am
Neighborhoods of a point include the point itself, so yes, f is differentiable at a.
Title: Re: Continuity & the derivative
Post by John_Gaughan on Apr 5th, 2004, 9:19pm
f(x) need not be continuous, but it only has a derivative where it is defined. Any discontinuities in f(x) result in similar discontinuities in dy/dx.

For example, if f(x) = |x|, then it has a derivative dy/dx except at x = 0. The derivative will be -1 when x < 0, 1 when x > 0, and not defined at 0.

Am I answering the correct question? This seems a bit basic.
Title: Re: Continuity & the derivative
Post by kellys on Apr 5th, 2004, 10:25pm

on 04/05/04 at 21:19:49, John_Gaughan wrote:
f(x) need not be continuous, but it only has a derivative where it is defined. Any discontinuities in f(x) result in similar discontinuities in dy/dx.

For example, if f(x) = |x|, then it has a derivative dy/dx except at x = 0. The derivative will be -1 when x < 0, 1 when x > 0, and not defined at 0.

Am I answering the correct question? This seems a bit basic.


The question is asking about the continuity of the derivative when f is differentiable.  The absolute value isn't differentiable at x=0.

I'm inclined to answer that the derivative must always be continuous.  By the definition of a limit, f is differentiable at a with derivative f'(a) iff its left and right derivatives exist and agree.  If you write out the limits, you get the definition of continuity for f' at a.
Title: Re: Continuity & the derivative
Post by Barukh on Apr 6th, 2004, 6:20am
This innocently looking question showed me once again how subtle are the concepts of calculus, and how little I know about analysis…  :( So, I consulted the books that I have in my possession (no textbook on analysis), revealing the following.

Instead of asking the question posted by Icarus, let's consider the question: given a function g(x) on an interval [b, c], what are the conditions that g is integrable on [b, c]? The extensive work of Riemann and Darboux showed that g(x) must satisfy the following conditions:

1. g(x) is bounded on [b, c].

2. The mean value theorem must hold, that is, if f(x) is defined as  [smiley=int.gif]bxg(x)dx, then f(c) – f(b) = g(d)(c-b), where d [in] [b, c].

2. The set of points { g(x) |  x [in] [b, c] } is itself an interval (for continuous function, this property is called an Intermediate Value Theorem).

3. The set of points where g(x) is discontinuous, has measure zero.

Because Darboux also showed that the fundamental theorem of the calculus - that differentiation and integration are inverse operations - holds also for the "extended case" just described, this also covers Icarus's question.

Does it make any sense? I would appreciate any remarks on incorrect/incomplete statements made. Are any particular examples available?
P.S. After rereading this, I understood that condition 2 doesn't make any sense, so I changed it to what I believe is correct. Sorry for inconvinence…
Title: Re: Continuity & the derivative
Post by Pietro K.C. on Apr 6th, 2004, 1:57pm
Wasn't it Lebesgue who introduced the concept of measure, much later? I think maybe what Riemann/Darboux showed was that the set of points in which g(x) can be discontinuous is denumerable.

Denumerable sets on the real line have measure zero, but I'm not sure whether the converse is true.
Title: Re: Continuity & the derivative
Post by Icarus on Apr 6th, 2004, 5:16pm

on 04/05/04 at 21:19:49, John_Gaughan wrote:
f(x) need not be continuous, but it only has a derivative where it is defined. Any discontinuities in f(x) result in similar discontinuities in dy/dx.


A basic result which should be included in any calculus book (other than baby .. erh.. business calculus) is that if f is differentiable at a, then f must also be continuous at a.

But as kellys pointed out, it is the continuity of f' = df/dx, not f, that I am asking about.


Quote:
For example, if f(x) = |x|, then it has a derivative dy/dx except at x = 0. The derivative will be -1 when x < 0, 1 when x > 0, and not defined at 0.


This does provide an example of why I ask the question. It certainly seems from this and other examples that when the derivative would be expected to discontinuous (such as here where it jumps from -1 to 1 at 0), it in fact fails to exist at all (|x| is not differentiable at 0).

Is this always true? Or are there examples of functions which are differentiable at 0 and around 0, but the derivative is not continuous at 0?


on 04/05/04 at 22:25:43, kellys wrote:
By the definition of a limit, f is differentiable at a with derivative f'(a) iff its left and right derivatives exist and agree.  If you write out the limits, you get the definition of continuity for f' at a.


Existance of the derivative:

limh[to]0+ (f(a+h) - f(a))/h = limh[to]0+ (f(a-h) - f(a))/h

Continuity of the derivative:

limx[to]a f'(x) = f(a).

I don't see how any restating of these gets one to look like the other.
Title: Re: Continuity & the derivative
Post by Icarus on Apr 6th, 2004, 5:31pm

on 04/06/04 at 06:20:13, Barukh wrote:
given a function g(x) on an interval [b, c], what are the conditions that g is integrable on [b, c]? The extensive work of Riemann and Darboux showed that g(x) must satisfy the following conditions:

1. g(x) is bounded on [b, c].

2. The mean value theorem must hold, that is, if f(x) is defined as  [smiley=int.gif]bxg(x)dx, then f(c) – f(b) = g(d)(c-b), where d [in] [b, c].

2. The set of points { g(x) |  x [in] [b, c] } is itself an interval (for continuous function, this property is called an Intermediate Value Theorem).

3. The set of points where g(x) is discontinuous, has measure zero.


This first begs the question, "integrable by which definition?". Since you bring up the idea of measure zero, one might assume Lesbegue, but then none of your claims is true. The function j(x) = 1/x if x is rational, j(x) = 0 if x is irrational, is discontinuous everywhere, is unbounded in any neighborhood of zero and image(j) = {0, 1}. Yet j is Lesbegue Integrable on any interval, and has integral 0.

If we restrict ourselves to Reimann integrability, (2) is still clearly false for any monotone function with an isolated jump discontinuity, yet such functions are easily seen to be Reimann Integrable. Condition (1) is usually a part of the definition of the Reimann integral, but later relaxed with the introduction of improper integrals. Certainly, f(x) = x-2, x [ne]0, f(0) = 0 is usually considered integrable on [0,1].
Title: Re: Continuity & the derivative
Post by kellys on Apr 6th, 2004, 5:34pm
I see what you're saying Icarus, I probably should have worked it out before I spoke.

If I were to guess, I would say that if such a function did exist, it would be a fractal.  For example, take the Koch curve and change the rules slightly so that it is a function over the reals.  I remember reading in some intro analysis notes that certain fractal are smooth, but by definition their derivative takes on finitely many values.
Title: Re: Continuity & the derivative
Post by Icarus on Apr 6th, 2004, 5:36pm

on 04/06/04 at 13:57:46, Pietro K.C. wrote:
Wasn't it Lebesgue who introduced the concept of measure, much later? I think maybe what Riemann/Darboux showed was that the set of points in which g(x) can be discontinuous is denumerable.

Denumerable sets on the real line have measure zero, but I'm not sure whether the converse is true.


The converse is not true. The Cantor set is not denumerable, but has measure zero.

I am not sure (my memory is not that strong), but I believe that Barukh is correct about the measure 0 requirement for Riemann integrability. It may even hold true for Lesbegue Integrability with a correction to the statement.
Title: Re: Continuity & the derivative
Post by Icarus on Apr 6th, 2004, 5:48pm

on 04/06/04 at 17:34:10, kellys wrote:
If I were to guess, I would say that if such a function did exist, it would be a fractal.  For example, take the Koch curve and change the rules slightly so that it is a function over the reals.  I remember reading in some intro analysis notes that certain fractal are smooth, but by definition their derivative takes on finitely many values.


I suppose that would depend exactly how you define "fractal". Weierstrass displayed a nowhere-differentiable continuous function nearly 100 years before Mandelbrot invented the term. One might call such functions "fractal" now, but they were well known long before Mandelbrot.
Title: Re: Continuity & the derivative
Post by Barukh on Apr 7th, 2004, 1:22am
Thanks for your comments, Icarus. As I said, I remember too little about calculus to get an intuition what is wrong and what is right… Therefore, what is left, is read the sources (more or less reliable), and take what they offer, if it doesn’t contradict the very basic knowledge. Of course, it doesn’t imply they are correct!


on 04/06/04 at 17:31:53, Icarus wrote:
This first begs the question, "integrable by which definition?".

You got it right: Riemann integral is considered. I learned that Darboux has extended the notion of the Riemann integral by introducing two sums, S and s, which are called the upper and lower integral respectively. The integral is defined if S = s. Then, he showed that for this “extended” case the fundamental theorem of the calculus still holds.

(BTW, does it hold if Lebesque integral is considered?).


Quote:
If we restrict ourselves to Reimann integrability, (2) is still clearly false for any monotone function with an isolated jump discontinuity, yet such functions are easily seen to be Reimann Integrable. Condition (1) is usually a part of the definition of the Reimann integral, but later relaxed with the introduction of improper integrals. Certainly, f(x) = x-2, x [ne]0, f(0) = 0 is usually considered integrable on [0,1].

The following page (http://planetmath.org/encyclopedia/IntermediateValuePropertyOfTheDerivative.html) actually states my second condition.
Title: Re: Continuity & the derivative
Post by Icarus on Apr 7th, 2004, 7:54pm
Yes, I had forgotten about Darboux's theorem. But you are talking here about a slightly different matter than "integrable".

To say that g is (Riemann) integrable on the interval [a, b] simply means that [int]ab g(x)dx exists. This does not mean that g has an antiderivative.

The Fundamental theorem of calculus says that if g is continuous, then G(x) = [int]ax g(t)dt is differentiable and G' = g.

If g is not continuous, it may still be integrable, but G' [ne] g in general.

Lesbegue integration extends Riemann integration in that some functions that are not Riemann integrable, are Lesbegue integrable. But every Riemann integrable function is also Lesbegue integrable, and the two integrals give the same value. That is, Lesbegue integration only extends Riemann integration - it does not change what was already there.

This means that the Fundamental theorem of Calculus holds just as well for Lesbegue integration as it does for Riemann.
Title: Re: Continuity & the derivative
Post by Barukh on Apr 11th, 2004, 10:25am
While searching the web for the subject, I’ve stumbled over  the Historical Essay On Continuity Of Derivatives (http://www.math.niu.edu/~rusin/known-math/00_incoming/deriv_cont) by Dave L. Renfro. I found it very interesting. One of the variants is called “Differentiable, Everywhere Oscillating Function”. The most general conditions for the derivative are stated in the first paragraph of the introduction. This – IMHO – answers the Icarus’s question.

BTW, anybody knows the meaning of the following terms: 1) Baire one function, 2) F_sigma set?
Title: Re: Continuity & the derivative
Post by Icarus on Apr 11th, 2004, 1:10pm
I haven't looked at your link yet, but wanted to answer your other questions.

An F-Sigma set is a countable union of closed sets. These sets come up occasionally in topology and analysis.

A Baire-1 function is a discontinuous function which is the pointwise limit of a sequence of continuous functions. More generally, Baire-0 functions are continuous functions, and Baire-k functions are functions which are not Baire-(k-1), but are the pointwise limits of sequences of Baire-(k-1) functions. This definition is new to me - I got it from Mathworld.
Title: Re: Continuity & the derivative
Post by Icarus on Apr 11th, 2004, 3:25pm
That link is interesting, but I had to go look up a lot of terms I'd forgotten over the years, and a few I'd never encountered!

To me, the most intriguing result was the one by Weil:

Let D = {f' | f is differentiable on all [bbr], and f' is bounded}. Then the subset G = { g [in] D | g is not discontinuous almost everywhere} is of the first category under the sup norm on D.

So the answer to my first question is no. And the second appears to have been the subject of much research.

My intent was not so deep, however. Can you provide an explicit example of a function that is differentiable at a point, but the derivative is not continuous there? I had a particular example in mind, but there are of course plenty of others.

Hint: [hide]Barukh's last link specifically mentioned "occilatory" functions.[/hide]
Title: Re: Continuity & the derivative
Post by Barukh on Apr 15th, 2004, 12:25am

on 04/11/04 at 15:25:02, Icarus wrote:
Can you provide an explicit example of a function that is differentiable at a point, but the derivative is not continuous there? I had a particular example in mind, but there are of course plenty of others.

The example I've found in literature is as follows: define F(x) = x2sin(1/x) if x [ne] 0; F(x) = 0 if x = 0. This function has a derivative = 0 at x = 0, but in every interval (0, [epsilon]) F’(x) takes on all the values in [-1, 1].


Quote:
My intent was not so deep, however.

I have photocopied the paper where a “simple” construction of “everywhere differentiable, nowhere monotone, function” is presented. Will see if I can manage to understand this one…


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