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        riddles >> putnam exam (pure math) >> Possible continuous/differentiable sets
        
(Message started by: Deedlit on May 4^th, 2005, 7:28pm)

Title: Possible continuous/differentiable sets
Post by Deedlit on May 4^th, 2005, 7:28pm

Here's a random question that I was curious about:

What are the possible subsets S of R such that, for some function f: R -> R, S is the set of points x such that:

1. f is continuous at x.
2. f is differentiable at x.
3. f is n times differentiable at x.
4. f is n times continuously differentiable at x.
5. f is infinitely differentiable at x.
6. f is analytic at x.

Okay, that's actually a bunch of questions. :D But I'd be interested in any ideas about any of the problems. I know the answer to 1 and 6, but not the others.

Title: Re: Possible continuous/differentiable sets
Post by Icarus on May 5^th, 2005, 6:39pm

I believe that the answer to (1) is all subsets of R, and I know that the answer to (6) is all open sets. The rest I am not sure about off the top of my head.

To see that (6) consists of open sets: simply note for f to be analytic at x, it must be representable as a power series in some neighborhood of x. And as such, it is analytic everywere in that neighborhood. So the set of all points where f is analytic is open. Contrawise, if S is any open set, define f(x) = e^x for x in S, and f(x) = 0 for x not in S.

Title: Re: Possible continuous/differentiable sets
Post by Deedlit on May 5^th, 2005, 10:16pm

For 1, can you define a function that is continuous on the rationals, but not the irrationals?

Yeah, that's what I got for 6, although you need to do more when x is not in S, of course. I wonder if we can handle every open set if the function must be C-infinity...

Title: Re: Possible continuous/differentiable sets
Post by Icarus on May 6^th, 2005, 7:39am

on 05/05/05 at 22:16:25, Deedlit wrote:

Yeah, that's what I got for 6, although you need to do more when x is not in S, of course.

Yeah - I must of been sleeping when I wrote that! To be more specific, we can set f equal to any non-analytic function for x not in S.

Quote:

I wonder if we can handle every open set if the function must be C-infinity...

I'm not sure I understand you here: analytic implies C^[infty], so the analytic case works for them. I suspect that you can get any F-Sigma or G-Delta set for the C^[infty] case.

Title: Re: Possible continuous/differentiable sets
Post by Deedlit on May 6^th, 2005, 7:55am

on 05/06/05 at 07:39:56, Icarus wrote:

Yeah - I must of been sleeping when I wrote that! To be more specific, we can set f equal to any non-analytic function for x not in S.

and also, never equal to e^x.

Quote:

I'm not sure I understand you here: analytic implies C^[infty], so the analytic case works for them. I suspect that you can get any F-Sigma or G-Delta set for the C^[infty] case.

Sorry, I was still talking about 6; Given an open set U, can we always find a C^[infty] everywhere function that is analytic precisely on U? For starters, can we define a C^[infty] function that is non-analytic everywhere?

Title: Re: Possible continuous/differentiable sets
Post by Barukh on May 6^th, 2005, 8:07am

on 05/05/05 at 22:16:25, Deedlit wrote:

For 1, can you define a function that is continuous on the rationals, but not the irrationals?

Interesting! It turns out that the set of discontinuities of any R -> R function is a countable union of closed sets. Because irrationals are not in this category, the function in question cannot be constructed.

This IMHO answers #1.

Title: Re: Possible continuous/differentiable sets
Post by Deedlit on May 6^th, 2005, 8:21am

Indeed! But, as usual, I must ask: Why?

Edit: I realize that, perhaps, you are holding back the solution to allow others a chance to solve it, as I have done myself. In that case, I don't mean to prod you into giving the answer prematurely. :)

Title: Re: Possible continuous/differentiable sets
Post by Icarus on May 7^th, 2005, 8:28am

on 05/06/05 at 07:55:51, Deedlit wrote:

and also, never equal to e^x.

It can be equal to e^x at individual points, just not on neighborhoods. But that is already implied by it not being analytic.

on 05/06/05 at 07:55:51, Deedlit wrote:

For starters, can we define a C^[infty] function that is non-analytic everywhere?

Let f(x) = e^(-x^-2) for x>0 and f(x) = 0 for x< 0. Let g(x) = f(x)/(f(x)+1). So g is C^[infty] everywhere, is analytic everywhere except at 0, and is bounded between 0 and 1.

Let {r_i} be an enumeration of the rationals, and define h(x) = [sum]_i 2^-ig(x - r_i). I believe h is C^[infty], but is nowhere analytic.

on 05/06/05 at 07:55:51, Deedlit wrote:

Given an open set U, can we always find a C^[infty] everywhere function that is analytic precisely on U?

Since any closed set in R is the union of a countable number of disjoint closed intervals, and given the existance of a smooth non-analytic function such as h, this is not hard.

First of all, define t(x) = f(x)f(1-x)h(x). t is zero everywhere except in the interval (0, 1). t is also not analytic anywhere in [0, 1]. Now, for each disjoint closed interval [a, b] in the compliment of U, define r(x) = t((x - a)/(b - a)). For x in U, define r(x) = 0. r is analytic exactly on U, but smooth everywhere.

Title: Re: Possible continuous/differentiable sets
Post by Deedlit on May 7^th, 2005, 4:45pm

on 05/07/05 at 08:28:58, Icarus wrote:

It can be equal to e^x at individual points, just not on neighborhoods. But that is already implied by it not being analytic.

For isolated points of R\S, though, you have to avoid matching up with the function on S. No biggie, of course!

Quote:

Yes, that looks like it can work. The problem is evaluating the derivatives - they can get quite nasty!

Quote:

Since any closed set in R is the union of a countable number of disjoint closed intervals,

Unfortunately, this isn't true. The Cantor set, for instance, has a continuum number of closed intervals, and they aren't isolated. Your construction doesn't seem to cover sets like this.

Title: Re: Possible continuous/differentiable sets
Post by Icarus on May 8^th, 2005, 11:58am

on 05/07/05 at 16:45:07, Deedlit wrote:

Yes, that looks like it can work. The problem is evaluating the derivatives - they can get quite nasty!

The derivatives are not a problem. Some well-known results will demonstrate that h is smooth (C-infinity). But demonstrating that h is not analytic is the part that stumps me. I am sure it is not, but don't see how to demonstrate it. It is conceivable that the non-analysity of g(x) at zero is lost with the sum.

Quote:

Unfortunately, this isn't true. The Cantor set, for instance, has a continuum number of closed intervals, and they aren't isolated. Your construction doesn't seem to cover sets like this.

Yeah. I seem to have flipped the result I was thinking of: any open set in R is the countable union of disjoint open intervals.

Title: Re: Possible continuous/differentiable sets
Post by Deedlit on May 21^st, 2005, 3:44am

This thread has gotten a bit old, but let me ask - what are the well-known results that prove that h is smooth?

Title: Re: Possible continuous/differentiable sets
Post by Eigenray on Jul 25^th, 2005, 5:47pm

on 05/06/05 at 08:07:36, Barukh wrote:

Interesting! It turns out that the set of discontinuities of any R -> R function is a countable union of closed sets.

Indeed. If U_n is the union of all open U with diam f(U)<1/n, then the intersection of the U_n is precisely the set of points of continuity of f.

Quote:

This IMHO answers #1.

Does anyone want to try the converse?

Quote:

Because irrationals are not in this category, the function in question cannot be constructed.

For, if Q=[cap]U_n, and we let V_n be the complement of the n-th rational, then each U_n and V_n are open and dense, while the intersection of all of them is empty, in contradiction to the Baire category theorem.

In a similar spirit: suppose f_n:R->R is a sequence of continuous functions converging pointwise to f. Show that f is continuous on an uncountable dense set.