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| Author |
Message |
atkrunner@hotmail.com
science forum beginner
Joined: 20 Jul 2006
Posts: 5
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 Posted: Thu Jul 20, 2006 12:46 am Post subject:
Function not in L_1 {[0,1]}, but satisfies ...?
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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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Ronald Bruck
science forum Guru
Joined: 05 Jun 2005
Posts: 356
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| Posted: Thu Jul 20, 2006 3:26 am Post subject:
Re: Function not in L_1 {[0,1]}, but satisfies ...?
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In article <1153356374.167733.130810@p79g2000cwp.googlegroups.com>,
<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 |
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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 ...?
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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 |
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Stephen Montgomery-Smith
science forum Guru
Joined: 01 May 2005
Posts: 487
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| Posted: Thu Jul 20, 2006 4:26 pm Post subject:
Re: Function not in L_1 {[0,1]}, but satisfies ...?
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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
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upendra0111@rediffmail.co
science forum beginner
Joined: 20 Jul 2006
Posts: 1
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| Posted: Thu Jul 20, 2006 4:48 pm Post subject:
Re: Function not in L_1 {[0,1]}, but satisfies ...?
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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:
| 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
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Stephen Montgomery-Smith
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 ...?
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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
|
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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 ...?
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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. |
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Stephen Montgomery-Smith
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 ...?
|
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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? |
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|
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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 ...?
|
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|
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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atkrunner@hotmail.com
science forum beginner
Joined: 20 Jul 2006
Posts: 5
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| Posted: Thu Jul 20, 2006 7:11 pm Post subject:
Re: Function not in L_1 {[0,1]}, but satisfies ...?
|
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|
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? |
|
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Stephen Montgomery-Smith
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)}. |
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Stephen Montgomery-Smith
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 } |
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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 ...?
|
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|
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 <DzOvg.1090104$xm3.733246@attbi_s21>, Stephen
Montgomery-Smith <stephen@math.missouri.edu> 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
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