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David C. Ullrich
science forum Guru

Joined: 28 Apr 2005
Posts: 2250

Posted: Thu Jul 13, 2006 10:35 am    Post subject: Re: Absolutely continuous, L^2 question

On Wed, 12 Jul 2006 15:39:57 EDT, James <james545@gmail.com> wrote:

 Quote: In article 5036829.1152728209310.JavaMail.jakarta@nitrogen.mathf orum.org>, James wrote: In article 1820258.1152720081570.JavaMail.jakarta@nitrogen.mathf orum.org>, James wrote: On Wed, 12 Jul 2006 09:00:56 EDT, James james545@gmail.com> wrote: This question has been killing me. Please help with any insight: Let f be absolutely continuous on [0,x] for all x 0. Let f, f ' be in L^2[0,oo). Let f(0) = 0. Prove that lim f(x) = 0 x --> oo There was a part (a) to this problem that I got : Prove that int_[0,x] |f f '| <= .5* (int_[0,x] |f '| )^2. But I don't see how this helps for part (b). I have tried several ways to prove that lim f(x) = 0 as x --> oo. 1) I have proven that if f is in L^1[0,oo) and f is uniformly continuous, then lim f(x) = 0 as x -- oo. Well, that fact that f' is in L^2 shows that f is uniformly continuous: |f(x+h) - f(x)| = ______ <= _______, which tends to 0 as h -> 0, uniformly in x. Are you asserting that f is uniformly continuous on all of [0,oo)? Your argument shows that f is uniformly continuous on [0,x] since f is AC on [0,x]. I don't think you know what his argument is. (Yes, f is UC on [0,oo).) Dear Wade, I don't see it. You say there is a universal delta on all of [0,oo)? |f(x+h) - f(x)| = |int_[x,x+h] f' | <= ||f'||_2 * h^(1/2) ---> 0 as h goes to 0. But to justify |f(x+h) - f(x)| = |int_[x,x+h] f' | you need to say that f is AC on [0,a] where a is greater than x+h. So you have that f is uniformly continuous on [0,a]. The fact that |f(x+h) - f(x)| --> 0 as h ---> 0 for all x in [0,a] doesn't give you what you want. If you want to say that f is uniformly continuous beyond a, then when you write |f(x+h) - f(x)| ---> 0 as h --> 0 (uniformly in x), this means that for all eps > 0 there is an s > 0 (i.e. delta > 0) with |f(x+h) - f(x)| < eps for all h < s for all x in [0,a]. But what I am saying is that if you want to check uniform continuity at y a, I think that this delta (or s) changes. In any case, whether or not my babbling above makes any sense, please share with me how you are getting your universal delta for the uniform continuity of f on all of[0,oo). If 0 <= x < y, you have |f(y) - f(x)| = |int_[x,y] f' | <= [int_[x,y] |f'|^2)^(1/2)]*|y-x|^(1/2) <= [int_[0,oo) |f'|^2)^(1/2)]*|y-x|^(1/2). So there is a constant C such that |f(y) - f(x)| <= C*|y-x|^(1/2) for all x, y in [0,oo). It's really the first part of your response that I am having trouble with. You say "If 0 <= x < y, you have |f(y) - f(x)| = |int_[x,y] f' | It seems to me that what you are basically saying is that f is absolutely continuous on all [0,oo). In order to say that |f(y) - f(x)| = |int_[x,y] f' |, you need to first say that f(y) = int_[0,y] f' and f(x) = int[0,x] f'. In order to say those two things you need to say that f is absolutely continuous on [0,z] where z is greater than y and x. So ok, given x and y, there is a z that makes this work.

Right. So what the heck is the problem?

 Quote: If you pick x' and y', then there is a z' that makes this work. So it changes.

So what? It's still true that for any x and y

f(y) - f(x) = int_x^y f'(t) dt,

because f is AC on [x,y]. How does the fact that if you change x and y
then you change x and y affect this?

 Quote: I am missing a logical step here. Side question : If f is absolutely continuous on [0,x] for all x > 0, then does this imply that f is absolutely continuous on [0,oo)?

No. And that doesn't matter one bit - all we use above is the fact
that it's AC on [x,y].

 Quote: James

************************

David C. Ullrich

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