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Deedlit
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Possible continuous/differentiable sets
« on: May 4th, 2005, 7:28pm » |
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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.  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.
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| « Last Edit: May 4th, 2005, 7:50pm by Deedlit » |
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Icarus
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Re: Possible continuous/differentiable sets
« Reply #1 on: May 5th, 2005, 6:39pm » |
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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) = ex for x in S, and f(x) = 0 for x not in S.
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Deedlit
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Re: Possible continuous/differentiable sets
« Reply #2 on: May 5th, 2005, 10:16pm » |
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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...
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Icarus
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Re: Possible continuous/differentiable sets
« Reply #3 on: May 6th, 2005, 7:39am » |
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on May 5th, 2005, 10:16pm, Deedlit wrote:| Yeah, that's what I got for 6, although you need to do more when x is not in S, of course. |
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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... |
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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.
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Deedlit
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Re: Possible continuous/differentiable sets
« Reply #4 on: May 6th, 2005, 7:55am » |
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on May 6th, 2005, 7:39am, 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.
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and also, never equal to ex.
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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. |
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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?
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Barukh
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Re: Possible continuous/differentiable sets
« Reply #5 on: May 6th, 2005, 8:07am » |
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on May 5th, 2005, 10:16pm, Deedlit wrote:| For 1, can you define a function that is continuous on the rationals, but not the irrationals? |
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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.
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Deedlit
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Re: Possible continuous/differentiable sets
« Reply #6 on: May 6th, 2005, 8:21am » |
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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. 
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| « Last Edit: May 6th, 2005, 8:29am by Deedlit » |
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Icarus
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Re: Possible continuous/differentiable sets
« Reply #7 on: May 7th, 2005, 8:28am » |
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on May 6th, 2005, 7:55am, Deedlit wrote:| and also, never equal to ex. |
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It can be equal to ex at individual points, just not on neighborhoods. But that is already implied by it not being analytic.
on May 6th, 2005, 7:55am, Deedlit wrote:| For starters, can we define a C[infty] function that is non-analytic everywhere? |
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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 {ri} be an enumeration of the rationals, and define h(x) = [sum]i 2-ig(x - ri). I believe h is C[infty], but is nowhere analytic.
on May 6th, 2005, 7:55am, Deedlit wrote:| Given an open set U, can we always find a C[infty] everywhere function that is analytic precisely on U? |
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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.
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Deedlit
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Re: Possible continuous/differentiable sets
« Reply #8 on: May 7th, 2005, 4:45pm » |
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on May 7th, 2005, 8:28am, Icarus wrote:
It can be equal to ex at individual points, just not on neighborhoods. But that is already implied by it not being analytic.
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For isolated points of R\S, though, you have to avoid matching up with the function on S. No biggie, of course!
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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 {ri} be an enumeration of the rationals, and define h(x) = [sum]i 2-ig(x - ri). I believe h is C[infty], but is nowhere analytic.
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Yes, that looks like it can work. The problem is evaluating the derivatives - they can get quite nasty!
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Since any closed set in R is the union of a countable number of disjoint closed intervals,
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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.
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| « Last Edit: May 7th, 2005, 4:45pm by Deedlit » |
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Icarus
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Re: Possible continuous/differentiable sets
« Reply #9 on: May 8th, 2005, 11:58am » |
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on May 7th, 2005, 4:45pm, Deedlit wrote:| Yes, that looks like it can work. The problem is evaluating the derivatives - they can get quite nasty! |
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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. |
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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.
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| « Last Edit: May 8th, 2005, 11:58am by Icarus » |
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Deedlit
Senior Riddler
Posts: 476
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Re: Possible continuous/differentiable sets
« Reply #10 on: May 21st, 2005, 3:44am » |
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This thread has gotten a bit old, but let me ask - what are the well-known results that prove that h is smooth?
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Eigenray
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Re: Possible continuous/differentiable sets
« Reply #11 on: Jul 25th, 2005, 5:47pm » |
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on May 6th, 2005, 8:07am, Barukh wrote:
Interesting! It turns out that the set of discontinuities of any R -> R function is a countable union of closed sets. |
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Indeed. If Un is the union of all open U with diam f(U)<1/n, then the intersection of the Un is precisely the set of points of continuity of f.
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Does anyone want to try the converse?
Quote:| Because irrationals are not in this category, the function in question cannot be constructed. |
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For, if Q=[cap]Un, and we let Vn be the complement of the n-th rational, then each Un and Vn 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 fn:R->R is a sequence of continuous functions converging pointwise to f. Show that f is continuous on an uncountable dense set.
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