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atkrunner@hotmail.com
science forum beginner

Joined: 20 Jul 2006
Posts: 5

Posted: Fri Jul 21, 2006 12:12 am    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...?

Okay I am obtuse, but I see it now. Your first response was right.
Just take the integral of the distribution function over [0, inf.),
which is infinite. And we know there is at least one function
with such a distribution.

I just talked with a classmate who says Tonelli is his
favorite trick in measure theory, now I see why.

Thanks both for the help.

John

Ronald Bruck wrote:
 Quote: Actually, that's why I gave the hint that f \in L^1 iff \sum_n n*m({x:n = f(x) < n+1}) is finite. That works out to approximately \sum_n (1/(n log n) - 1/((n+1)log(n+1))) \approx \sum_n n D_n (-1/(n log n)) \approx \sum_n 1/(n (log n)^2) + 1/(n log n) which is infinite. But I like your way much better. I didn't spend much time on it, because it's obvious, as you say. I guess f = g^{-1} isn't explicit enough for him. I suppose we could take something asymptotic to this which is close enough to work, and simple in form, but I'm too lazy to try. (And it's NOT NECESSARY.) -- Ron Bruck In article , Stephen Montgomery-Smith wrote: That's what we are doing, only we are giving hints instead of telling you explicitly. Remember f is not in L^1 if and only if ||f||_1=infty. upendra0111@rediffmail.com wrote: No, that is not what I am looking for. What I am asking is the following: Find a measurable function f, on [0,1] such that 1) f is not in L^1, and 2) r* m{f>=r} ->0 as r ->infinity. Stephen Montgomery-Smith wrote: atkrunner@hotmail.com wrote: My wording may have been a bit unclear. Yes, I did want the function to be nonnegative. And such a function will need to have a distribution function of the type you mentioned, namely r*g(r) -->0 as r-->infinity. But I am looking for an explicit example of such a function, defined on the unit interval. Perhaps it is decreasing. If so, and we take g(r) as you described, then f can be defined as the inverse of 1/{(r+1)*log(r+1)}, restricted to [0,1]. Can you show that this function is not in L^1{[0,1]}? If f is positive, then ||f||_1 = int_0^infty m(f>t) dt. (The easiest 'proof' is to draw a picture.) Thanks, John Ronald Bruck wrote: In article <1153356374.167733.130810@p79g2000cwp.googlegroups.com>, atkrunner@hotmail.com> wrote: Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that r* m( f >= r) --> 0 as r --> infinity? Where m is Lebesgue measure. If we assume f is mon. decreasing, then the condition states that N* f^-1(N) -->0 as N --> inf. If f is in L_1, then the above limit holds. So I am really trying to show the converse is false, but I have been unable to come up with a counter example. Isn't the counterexample intrinsic to the problem? First change your measure space to [0,infinity) [easy], then let g denote the distribution function of f, g(r) = m({x : f(x) >= r}). I'm going to assume you want f >= 0 (else this is trivial). Just take g to be anything which works, i.e. it's measurable and decreases to 0, e.g. g(r) = 1/((r+1)log(r+1)), then choose an f which has g as its distribution function. We have r g(r) --> 0 as r --> infinity, f >= 0, and f is not in L^1. Am I missing something? If f IS in L^1, then \sum_n n * m({x : n <= f(x) < n+1)} is convergent (an easy exercise on partial summation). And \int f dm = \int_0^\infty r dg(r), where dg(r) is the Lebesgue-Stieltjes measure induced by g. Or maybe that's a minus sign, since g is decreasing here... -- Ron Bruck Posted Via Usenet.com Premium Usenet Newsgroup Services ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com
Stephen Montgomery-Smith1
science forum Guru

Joined: 01 May 2005
Posts: 487

Posted: Thu Jul 20, 2006 7:39 pm    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...?

atkrunner@hotmail.com wrote:
 Quote: Just as I said before, m{f>=r}=1/{(r+1)*log(r+1)}, and Ron already mentioned that r* m{f>=r} ->0 in the case. This is fine. As far as ||f||_1, where we have restricted f to [0,1]?, I am stumped. Any help toward this end will be greatly appreciated. The inverse of 1/{(r+1)*log(r+1)}, restricted to [0,1]. Hint, what is m{f>=r}? Then what is ||f||_1?

Or, notice that if f is restricted to [0,1]

m{f>=r} = min{ 1/{(r+1)*log(r+1)} , 1 }
Stephen Montgomery-Smith1
science forum Guru

Joined: 01 May 2005
Posts: 487

Posted: Thu Jul 20, 2006 7:33 pm    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...?

atkrunner@hotmail.com wrote:
 Quote: Just as I said before, m{f>=r}=1/{(r+1)*log(r+1)}, and Ron already mentioned that r* m{f>=r} ->0 in the case. This is fine. As far as ||f||_1, where we have restricted f to [0,1]?, I am stumped. Any help toward this end will be greatly appreciated. The inverse of 1/{(r+1)*log(r+1)}, restricted to [0,1]. Hint, what is m{f>=r}? Then what is ||f||_1?

OK. Take f such that m{f>=r}=1/{(r+1)*log(r+e)}.
atkrunner@hotmail.com
science forum beginner

Joined: 20 Jul 2006
Posts: 5

Posted: Thu Jul 20, 2006 7:11 pm    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...?

Just as I said before, m{f>=r}=1/{(r+1)*log(r+1)}, and
Ron already mentioned that r* m{f>=r} ->0 in the case.
This is fine.

As far as ||f||_1, where we have restricted f to [0,1]?,
I am stumped.

Any help toward this end will be greatly appreciated.

 Quote: The inverse of 1/{(r+1)*log(r+1)}, restricted to [0,1]. Hint, what is m{f>=r}? Then what is ||f||_1?
Ronald Bruck
science forum Guru

Joined: 05 Jun 2005
Posts: 356

Posted: Thu Jul 20, 2006 7:10 pm    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...?

Actually, that's why I gave the hint that f \in L^1 iff \sum_n n*m({x:n
<= f(x) < n+1}) is finite. That works out to approximately

\sum_n (1/(n log n) - 1/((n+1)log(n+1)))

\approx \sum_n n D_n (-1/(n log n))

\approx \sum_n 1/(n (log n)^2) + 1/(n log n)

which is infinite. But I like your way much better. I didn't spend
much time on it, because it's obvious, as you say.

I guess f = g^{-1} isn't explicit enough for him. I suppose we could
take something asymptotic to this which is close enough to work, and
simple in form, but I'm too lazy to try. (And it's NOT NECESSARY.)

--
Ron Bruck

In article <DzOvg.1090104\$xm3.733246@attbi_s21>, Stephen
Montgomery-Smith <stephen@math.missouri.edu> wrote:

 Quote: That's what we are doing, only we are giving hints instead of telling you explicitly. Remember f is not in L^1 if and only if ||f||_1=infty. upendra0111@rediffmail.com wrote: No, that is not what I am looking for. What I am asking is the following: Find a measurable function f, on [0,1] such that 1) f is not in L^1, and 2) r* m{f>=r} ->0 as r ->infinity. Stephen Montgomery-Smith wrote: atkrunner@hotmail.com wrote: My wording may have been a bit unclear. Yes, I did want the function to be nonnegative. And such a function will need to have a distribution function of the type you mentioned, namely r*g(r) -->0 as r-->infinity. But I am looking for an explicit example of such a function, defined on the unit interval. Perhaps it is decreasing. If so, and we take g(r) as you described, then f can be defined as the inverse of 1/{(r+1)*log(r+1)}, restricted to [0,1]. Can you show that this function is not in L^1{[0,1]}? If f is positive, then ||f||_1 = int_0^infty m(f>t) dt. (The easiest 'proof' is to draw a picture.) Thanks, John Ronald Bruck wrote: In article <1153356374.167733.130810@p79g2000cwp.googlegroups.com>, atkrunner@hotmail.com> wrote: Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that r* m( f >= r) --> 0 as r --> infinity? Where m is Lebesgue measure. If we assume f is mon. decreasing, then the condition states that N* f^-1(N) -->0 as N --> inf. If f is in L_1, then the above limit holds. So I am really trying to show the converse is false, but I have been unable to come up with a counter example. Isn't the counterexample intrinsic to the problem? First change your measure space to [0,infinity) [easy], then let g denote the distribution function of f, g(r) = m({x : f(x) >= r}). I'm going to assume you want f >= 0 (else this is trivial). Just take g to be anything which works, i.e. it's measurable and decreases to 0, e.g. g(r) = 1/((r+1)log(r+1)), then choose an f which has g as its distribution function. We have r g(r) --> 0 as r --> infinity, f >= 0, and f is not in L^1. Am I missing something? If f IS in L^1, then \sum_n n * m({x : n <= f(x) < n+1)} is convergent (an easy exercise on partial summation). And \int f dm = \int_0^\infty r dg(r), where dg(r) is the Lebesgue-Stieltjes measure induced by g. Or maybe that's a minus sign, since g is decreasing here... -- Ron Bruck

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Stephen Montgomery-Smith1
science forum Guru

Joined: 01 May 2005
Posts: 487

Posted: Thu Jul 20, 2006 6:46 pm    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...?

atkrunner@hotmail.com wrote:
 Quote: Stephen Montgomery-Smith wrote: That's what we are doing, only we are giving hints instead of telling you explicitly. Remember f is not in L^1 if and only if ||f||_1=infty. I know, that is just the definition of L^1. If f is positive, then ||f||_1 = int_0^infty m(f>t) dt. (The easiest 'proof' is to draw a picture.) I can show this statement also. All of the statements you both have offered are true, and I can prove them. I am actually not trying to prove anything, however. I am only looking for a specific function that has the following properties: Find a measurable function f, on [0,1] such that 1) f is not in L^1, and 2) r* m{f>=r} ->0 as r ->infinity. Can someone find such a function? If so, please let me know.

The inverse of 1/{(r+1)*log(r+1)}, restricted to [0,1]. Hint, what is
m{f>=r}? Then what is ||f||_1?
atkrunner@hotmail.com
science forum beginner

Joined: 20 Jul 2006
Posts: 5

Posted: Thu Jul 20, 2006 5:36 pm    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...?

Stephen Montgomery-Smith wrote:
 Quote: That's what we are doing, only we are giving hints instead of telling you explicitly. Remember f is not in L^1 if and only if ||f||_1=infty.

I know, that is just the definition of L^1.

 Quote: If f is positive, then ||f||_1 = int_0^infty m(f>t) dt. (The easiest 'proof' is to draw a picture.)

I can show this statement also.

All of the statements you both have offered
are true, and I can prove them. I am actually
not trying to prove anything, however. I am only
looking for a specific function that has the following
properties:

Find a measurable function f, on [0,1] such that

1) f is not in L^1, and

2) r* m{f>=r} ->0 as r ->infinity.

Can someone find such a function? If so, please let
me know.
Stephen Montgomery-Smith1
science forum Guru

Joined: 01 May 2005
Posts: 487

Posted: Thu Jul 20, 2006 4:54 pm    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...?

That's what we are doing, only we are giving hints instead of telling
you explicitly. Remember f is not in L^1 if and only if ||f||_1=infty.

upendra0111@rediffmail.com wrote:
 Quote: No, that is not what I am looking for. What I am asking is the following: Find a measurable function f, on [0,1] such that 1) f is not in L^1, and 2) r* m{f>=r} ->0 as r ->infinity. Stephen Montgomery-Smith wrote: atkrunner@hotmail.com wrote: My wording may have been a bit unclear. Yes, I did want the function to be nonnegative. And such a function will need to have a distribution function of the type you mentioned, namely r*g(r) -->0 as r-->infinity. But I am looking for an explicit example of such a function, defined on the unit interval. Perhaps it is decreasing. If so, and we take g(r) as you described, then f can be defined as the inverse of 1/{(r+1)*log(r+1)}, restricted to [0,1]. Can you show that this function is not in L^1{[0,1]}? If f is positive, then ||f||_1 = int_0^infty m(f>t) dt. (The easiest 'proof' is to draw a picture.) Thanks, John Ronald Bruck wrote: In article <1153356374.167733.130810@p79g2000cwp.googlegroups.com>, atkrunner@hotmail.com> wrote: Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that r* m( f >= r) --> 0 as r --> infinity? Where m is Lebesgue measure. If we assume f is mon. decreasing, then the condition states that N* f^-1(N) -->0 as N --> inf. If f is in L_1, then the above limit holds. So I am really trying to show the converse is false, but I have been unable to come up with a counter example. Isn't the counterexample intrinsic to the problem? First change your measure space to [0,infinity) [easy], then let g denote the distribution function of f, g(r) = m({x : f(x) >= r}). I'm going to assume you want f >= 0 (else this is trivial). Just take g to be anything which works, i.e. it's measurable and decreases to 0, e.g. g(r) = 1/((r+1)log(r+1)), then choose an f which has g as its distribution function. We have r g(r) --> 0 as r --> infinity, f >= 0, and f is not in L^1. Am I missing something? If f IS in L^1, then \sum_n n * m({x : n <= f(x) < n+1)} is convergent (an easy exercise on partial summation). And \int f dm = \int_0^\infty r dg(r), where dg(r) is the Lebesgue-Stieltjes measure induced by g. Or maybe that's a minus sign, since g is decreasing here... -- Ron Bruck
upendra0111@rediffmail.co
science forum beginner

Joined: 20 Jul 2006
Posts: 1

Posted: Thu Jul 20, 2006 4:48 pm    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...?

No, that is not what I am looking for. What I am

Find a measurable function f, on [0,1] such that

1) f is not in L^1, and

2) r* m{f>=r} ->0 as r ->infinity.

Stephen Montgomery-Smith wrote:
 Quote: atkrunner@hotmail.com wrote: My wording may have been a bit unclear. Yes, I did want the function to be nonnegative. And such a function will need to have a distribution function of the type you mentioned, namely r*g(r) -->0 as r-->infinity. But I am looking for an explicit example of such a function, defined on the unit interval. Perhaps it is decreasing. If so, and we take g(r) as you described, then f can be defined as the inverse of 1/{(r+1)*log(r+1)}, restricted to [0,1]. Can you show that this function is not in L^1{[0,1]}? If f is positive, then ||f||_1 = int_0^infty m(f>t) dt. (The easiest 'proof' is to draw a picture.) Thanks, John Ronald Bruck wrote: In article <1153356374.167733.130810@p79g2000cwp.googlegroups.com>, atkrunner@hotmail.com> wrote: Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that r* m( f >= r) --> 0 as r --> infinity? Where m is Lebesgue measure. If we assume f is mon. decreasing, then the condition states that N* f^-1(N) -->0 as N --> inf. If f is in L_1, then the above limit holds. So I am really trying to show the converse is false, but I have been unable to come up with a counter example. Isn't the counterexample intrinsic to the problem? First change your measure space to [0,infinity) [easy], then let g denote the distribution function of f, g(r) = m({x : f(x) >= r}). I'm going to assume you want f >= 0 (else this is trivial). Just take g to be anything which works, i.e. it's measurable and decreases to 0, e.g. g(r) = 1/((r+1)log(r+1)), then choose an f which has g as its distribution function. We have r g(r) --> 0 as r --> infinity, f >= 0, and f is not in L^1. Am I missing something? If f IS in L^1, then \sum_n n * m({x : n <= f(x) < n+1)} is convergent (an easy exercise on partial summation). And \int f dm = \int_0^\infty r dg(r), where dg(r) is the Lebesgue-Stieltjes measure induced by g. Or maybe that's a minus sign, since g is decreasing here... -- Ron Bruck
Stephen Montgomery-Smith1
science forum Guru

Joined: 01 May 2005
Posts: 487

Posted: Thu Jul 20, 2006 4:26 pm    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...?

atkrunner@hotmail.com wrote:
 Quote: My wording may have been a bit unclear. Yes, I did want the function to be nonnegative. And such a function will need to have a distribution function of the type you mentioned, namely r*g(r) -->0 as r-->infinity. But I am looking for an explicit example of such a function, defined on the unit interval. Perhaps it is decreasing. If so, and we take g(r) as you described, then f can be defined as the inverse of 1/{(r+1)*log(r+1)}, restricted to [0,1]. Can you show that this function is not in L^1{[0,1]}?

If f is positive, then ||f||_1 = int_0^infty m(f>t) dt. (The easiest
'proof' is to draw a picture.)

 Quote: Thanks, John Ronald Bruck wrote: In article <1153356374.167733.130810@p79g2000cwp.googlegroups.com>, atkrunner@hotmail.com> wrote: Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that r* m( f >= r) --> 0 as r --> infinity? Where m is Lebesgue measure. If we assume f is mon. decreasing, then the condition states that N* f^-1(N) -->0 as N --> inf. If f is in L_1, then the above limit holds. So I am really trying to show the converse is false, but I have been unable to come up with a counter example. Isn't the counterexample intrinsic to the problem? First change your measure space to [0,infinity) [easy], then let g denote the distribution function of f, g(r) = m({x : f(x) >= r}). I'm going to assume you want f >= 0 (else this is trivial). Just take g to be anything which works, i.e. it's measurable and decreases to 0, e.g. g(r) = 1/((r+1)log(r+1)), then choose an f which has g as its distribution function. We have r g(r) --> 0 as r --> infinity, f >= 0, and f is not in L^1. Am I missing something? If f IS in L^1, then \sum_n n * m({x : n <= f(x) < n+1)} is convergent (an easy exercise on partial summation). And \int f dm = \int_0^\infty r dg(r), where dg(r) is the Lebesgue-Stieltjes measure induced by g. Or maybe that's a minus sign, since g is decreasing here... -- Ron Bruck
atkrunner@hotmail.com
science forum beginner

Joined: 20 Jul 2006
Posts: 5

Posted: Thu Jul 20, 2006 4:43 am    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...?

My wording may have been a bit unclear.

Yes, I did want the function to be nonnegative. And such a function
will
need to have a distribution function of the type you mentioned, namely
r*g(r) -->0 as r-->infinity.

But I am looking for an explicit example of such a function, defined on
the unit interval. Perhaps it is decreasing. If so, and we take g(r)
as
you described, then f can be defined as the inverse of
1/{(r+1)*log(r+1)},
restricted to [0,1].

Can you show that this function is not in L^1{[0,1]}?

Thanks,
John

Ronald Bruck wrote:
 Quote: In article <1153356374.167733.130810@p79g2000cwp.googlegroups.com>, atkrunner@hotmail.com> wrote: Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that r* m( f >= r) --> 0 as r --> infinity? Where m is Lebesgue measure. If we assume f is mon. decreasing, then the condition states that N* f^-1(N) -->0 as N --> inf. If f is in L_1, then the above limit holds. So I am really trying to show the converse is false, but I have been unable to come up with a counter example. Isn't the counterexample intrinsic to the problem? First change your measure space to [0,infinity) [easy], then let g denote the distribution function of f, g(r) = m({x : f(x) >= r}). I'm going to assume you want f >= 0 (else this is trivial). Just take g to be anything which works, i.e. it's measurable and decreases to 0, e.g. g(r) = 1/((r+1)log(r+1)), then choose an f which has g as its distribution function. We have r g(r) --> 0 as r --> infinity, f >= 0, and f is not in L^1. Am I missing something? If f IS in L^1, then \sum_n n * m({x : n <= f(x) < n+1)} is convergent (an easy exercise on partial summation). And \int f dm = \int_0^\infty r dg(r), where dg(r) is the Lebesgue-Stieltjes measure induced by g. Or maybe that's a minus sign, since g is decreasing here... -- Ron Bruck
Ronald Bruck
science forum Guru

Joined: 05 Jun 2005
Posts: 356

Posted: Thu Jul 20, 2006 3:26 am    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...?

<atkrunner@hotmail.com> wrote:

 Quote: Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that r* m( f >= r) --> 0 as r --> infinity? Where m is Lebesgue measure. If we assume f is mon. decreasing, then the condition states that N* f^-1(N) -->0 as N --> inf. If f is in L_1, then the above limit holds. So I am really trying to show the converse is false, but I have been unable to come up with a counter example.

Isn't the counterexample intrinsic to the problem?

First change your measure space to [0,infinity) [easy], then let g
denote the distribution function of f,

g(r) = m({x : f(x) >= r}).

I'm going to assume you want f >= 0 (else this is trivial). Just take
g to be anything which works, i.e. it's measurable and decreases to 0,
e.g.

g(r) = 1/((r+1)log(r+1)),

then choose an f which has g as its distribution function. We have
r g(r) --> 0 as r --> infinity, f >= 0, and f is not in L^1.

Am I missing something?

If f IS in L^1, then \sum_n n * m({x : n <= f(x) < n+1)} is convergent
(an easy exercise on partial summation). And

\int f dm = \int_0^\infty r dg(r),

where dg(r) is the Lebesgue-Stieltjes measure induced by g. Or maybe
that's a minus sign, since g is decreasing here...

--
Ron Bruck
atkrunner@hotmail.com
science forum beginner

Joined: 20 Jul 2006
Posts: 5

 Posted: Thu Jul 20, 2006 12:46 am    Post subject: Function not in L_1 {[0,1]}, but satisfies ...? Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that r* m( f >= r) --> 0 as r --> infinity? Where m is Lebesgue measure. If we assume f is mon. decreasing, then the condition states that N* f^-1(N) -->0 as N --> inf. If f is in L_1, then the above limit holds. So I am really trying to show the converse is false, but I have been unable to come up with a counter example. thanks...

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