$frac{1}{||x||^n}$ is integrable on complement of $B(0,a) subset mathbb{R}^n$
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I am trying to prove that for a measurable function $f$, the Hardy-Littlewood maximal function is not in $L^1$ unless $f$ is $0$ a.e.
I was able to udnerstand the proof of the inequality here with $1$ replaced by an arbitrary constant $a>0$ : A question about the Hardy-Littlewood maximal function.
However I want to conclude by saying that Mf is not integrable since $1/|x|^n$ is not integrable on $B(0,a)^c$. Is this true? Am I thinking in the right direction?
EDIT: I found this in Frank Jones' measure theory book: https://imgur.com/a/tct1vI2. So it seems to be true, I just have no clue why it should be true.
real-analysis measure-theory
asked Nov 17 at 8:50
HerrWarum
15911
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up vote
0
down vote
favorite
I am trying to prove that for a measurable function $f$, the Hardy-Littlewood maximal function is not in $L^1$ unless $f$ is $0$ a.e.
I was able to udnerstand the proof of the inequality here with $1$ replaced by an arbitrary constant $a>0$ : A question about the Hardy-Littlewood maximal function.
However I want to conclude by saying that Mf is not integrable since $1/|x|^n$ is not integrable on $B(0,a)^c$. Is this true? Am I thinking in the right direction?
EDIT: I found this in Frank Jones' measure theory book: https://imgur.com/a/tct1vI2. So it seems to be true, I just have no clue why it should be true.
real-analysis measure-theory
asked Nov 17 at 8:50
HerrWarum
15911
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to prove that for a measurable function $f$, the Hardy-Littlewood maximal function is not in $L^1$ unless $f$ is $0$ a.e.
I was able to udnerstand the proof of the inequality here with $1$ replaced by an arbitrary constant $a>0$ : A question about the Hardy-Littlewood maximal function.
However I want to conclude by saying that Mf is not integrable since $1/|x|^n$ is not integrable on $B(0,a)^c$. Is this true? Am I thinking in the right direction?
EDIT: I found this in Frank Jones' measure theory book: https://imgur.com/a/tct1vI2. So it seems to be true, I just have no clue why it should be true.
real-analysis measure-theory
asked Nov 17 at 8:50
HerrWarum
15911
I am trying to prove that for a measurable function $f$, the Hardy-Littlewood maximal function is not in $L^1$ unless $f$ is $0$ a.e.
I was able to udnerstand the proof of the inequality here with $1$ replaced by an arbitrary constant $a>0$ : A question about the Hardy-Littlewood maximal function.
However I want to conclude by saying that Mf is not integrable since $1/|x|^n$ is not integrable on $B(0,a)^c$. Is this true? Am I thinking in the right direction?
EDIT: I found this in Frank Jones' measure theory book: https://imgur.com/a/tct1vI2. So it seems to be true, I just have no clue why it should be true.
real-analysis measure-theory
real-analysis measure-theory
asked Nov 17 at 8:50
HerrWarum
15911
asked Nov 17 at 8:50
HerrWarum
15911
edited Nov 17 at 8:57
asked Nov 17 at 8:50
HerrWarum
15911
asked Nov 17 at 8:50
HerrWarum
15911
asked Nov 17 at 8:50
HerrWarum
15911
15911
add a comment |
add a comment |
1 Answer
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It is true and it follows immediately if you use polar coordinates in $mathbb R^{n}$. Look for 'polar coordinates' in the index of Rudin's RCA.
answered Nov 17 at 12:01
Kavi Rama Murthy
43k31751
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
It is true and it follows immediately if you use polar coordinates in $mathbb R^{n}$. Look for 'polar coordinates' in the index of Rudin's RCA.
answered Nov 17 at 12:01
Kavi Rama Murthy
43k31751
add a comment |
up vote
0
down vote
It is true and it follows immediately if you use polar coordinates in $mathbb R^{n}$. Look for 'polar coordinates' in the index of Rudin's RCA.
answered Nov 17 at 12:01
Kavi Rama Murthy
43k31751
add a comment |
up vote
0
down vote
up vote
0
down vote
It is true and it follows immediately if you use polar coordinates in $mathbb R^{n}$. Look for 'polar coordinates' in the index of Rudin's RCA.
answered Nov 17 at 12:01
Kavi Rama Murthy
43k31751
It is true and it follows immediately if you use polar coordinates in $mathbb R^{n}$. Look for 'polar coordinates' in the index of Rudin's RCA.
answered Nov 17 at 12:01
Kavi Rama Murthy
43k31751
answered Nov 17 at 12:01
Kavi Rama Murthy
43k31751
answered Nov 17 at 12:01
Kavi Rama Murthy
43k31751
answered Nov 17 at 12:01
Kavi Rama Murthy
43k31751
43k31751
add a comment |
add a comment |
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