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 <james545@gmail.com> wrote:
In article
1820258.1152720081570.JavaMail.jakarta@nitrogen.mathf
orum.org>,
James <james545@gmail.com> 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)]*yx^(1/2) <= [int_[0,oo)
f'^2)^(1/2)]*yx^(1/2). So there is a constant C
such that
f(y)  f(x) <= C*yx^(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].
************************
David C. Ullrich 
