Mixing
Why does the lower bound on the Hardy-Littlewood maximal function make it non-integrable?
We have that $$Hf(x) geq frac{c}{|x|^n}$$ for some $c>0$ whenever $|x| geq 1$. How does this lower bound show that that the maximal function is non-integrable? Perhaps if we could show that $frac{c}{|x|^n}$ isn't integrable outside the unit ball we could show this. However, I am not sure how to do this. Similar questions seem to imply that it is obvious given the inequality.
measure-theory lebesgue-integral lebesgue-measure
asked Nov 21 '18 at 5:41
Jabbath
545
add a comment |
We have that $$Hf(x) geq frac{c}{|x|^n}$$ for some $c>0$ whenever $|x| geq 1$. How does this lower bound show that that the maximal function is non-integrable? Perhaps if we could show that $frac{c}{|x|^n}$ isn't integrable outside the unit ball we could show this. However, I am not sure how to do this. Similar questions seem to imply that it is obvious given the inequality.
measure-theory lebesgue-integral lebesgue-measure
asked Nov 21 '18 at 5:41
Jabbath
545
Use polar coordinates.
– Kavi Rama Murthy
Nov 21 '18 at 5:45
I'm not sure how to use polar coordinates in $R^n$. Is there a way to do it without polar coordinates?
– Jabbath
Nov 21 '18 at 6:09
1
Rudin's RCA tells you how to use polar coordinates in $mathbb R^{n}$.
– Kavi Rama Murthy
Nov 21 '18 at 7:48
add a comment |
We have that $$Hf(x) geq frac{c}{|x|^n}$$ for some $c>0$ whenever $|x| geq 1$. How does this lower bound show that that the maximal function is non-integrable? Perhaps if we could show that $frac{c}{|x|^n}$ isn't integrable outside the unit ball we could show this. However, I am not sure how to do this. Similar questions seem to imply that it is obvious given the inequality.
measure-theory lebesgue-integral lebesgue-measure
asked Nov 21 '18 at 5:41
Jabbath
545
We have that $$Hf(x) geq frac{c}{|x|^n}$$ for some $c>0$ whenever $|x| geq 1$. How does this lower bound show that that the maximal function is non-integrable? Perhaps if we could show that $frac{c}{|x|^n}$ isn't integrable outside the unit ball we could show this. However, I am not sure how to do this. Similar questions seem to imply that it is obvious given the inequality.
measure-theory lebesgue-integral lebesgue-measure
measure-theory lebesgue-integral lebesgue-measure
asked Nov 21 '18 at 5:41
Jabbath
545
asked Nov 21 '18 at 5:41
Jabbath
545
asked Nov 21 '18 at 5:41
Jabbath
545
asked Nov 21 '18 at 5:41
Jabbath
545
asked Nov 21 '18 at 5:41
Jabbath
545
545
Use polar coordinates.
– Kavi Rama Murthy
Nov 21 '18 at 5:45
I'm not sure how to use polar coordinates in $R^n$. Is there a way to do it without polar coordinates?
– Jabbath
Nov 21 '18 at 6:09
1
Rudin's RCA tells you how to use polar coordinates in $mathbb R^{n}$.
– Kavi Rama Murthy
Nov 21 '18 at 7:48
add a comment |
Use polar coordinates.
– Kavi Rama Murthy
Nov 21 '18 at 5:45
I'm not sure how to use polar coordinates in $R^n$. Is there a way to do it without polar coordinates?
– Jabbath
Nov 21 '18 at 6:09
1
Rudin's RCA tells you how to use polar coordinates in $mathbb R^{n}$.
– Kavi Rama Murthy
Nov 21 '18 at 7:48
Use polar coordinates.
– Kavi Rama Murthy
Nov 21 '18 at 5:45
Use polar coordinates.
– Kavi Rama Murthy
Nov 21 '18 at 5:45
I'm not sure how to use polar coordinates in $R^n$. Is there a way to do it without polar coordinates?
– Jabbath
Nov 21 '18 at 6:09
I'm not sure how to use polar coordinates in $R^n$. Is there a way to do it without polar coordinates?
– Jabbath
Nov 21 '18 at 6:09
1
1
Rudin's RCA tells you how to use polar coordinates in $mathbb R^{n}$.
– Kavi Rama Murthy
Nov 21 '18 at 7:48
Rudin's RCA tells you how to use polar coordinates in $mathbb R^{n}$.
– Kavi Rama Murthy
Nov 21 '18 at 7:48
add a comment |
1 Answer
1
active
oldest
votes
Hint: Observe
begin{align}
int_{mathbb{R}^n} frac{dx}{|x|^n} =int^infty_0 int_{|x|=r} frac{1}{|x|^n} dS(x)dr
end{align}
answered Nov 21 '18 at 7:49
Jacky Chong
17.8k21128
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
Hint: Observe
begin{align}
int_{mathbb{R}^n} frac{dx}{|x|^n} =int^infty_0 int_{|x|=r} frac{1}{|x|^n} dS(x)dr
end{align}
answered Nov 21 '18 at 7:49
Jacky Chong
17.8k21128
add a comment |
Hint: Observe
begin{align}
int_{mathbb{R}^n} frac{dx}{|x|^n} =int^infty_0 int_{|x|=r} frac{1}{|x|^n} dS(x)dr
end{align}
answered Nov 21 '18 at 7:49
Jacky Chong
17.8k21128
add a comment |
Hint: Observe
begin{align}
int_{mathbb{R}^n} frac{dx}{|x|^n} =int^infty_0 int_{|x|=r} frac{1}{|x|^n} dS(x)dr
end{align}
answered Nov 21 '18 at 7:49
Jacky Chong
17.8k21128
Hint: Observe
begin{align}
int_{mathbb{R}^n} frac{dx}{|x|^n} =int^infty_0 int_{|x|=r} frac{1}{|x|^n} dS(x)dr
end{align}
answered Nov 21 '18 at 7:49
Jacky Chong
17.8k21128
answered Nov 21 '18 at 7:49
Jacky Chong
17.8k21128
answered Nov 21 '18 at 7:49
Jacky Chong
17.8k21128
answered Nov 21 '18 at 7:49
Jacky Chong
17.8k21128
17.8k21128
add a comment |
add a comment |
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Use polar coordinates.
– Kavi Rama Murthy
Nov 21 '18 at 5:45
I'm not sure how to use polar coordinates in $R^n$. Is there a way to do it without polar coordinates?
– Jabbath
Nov 21 '18 at 6:09
1
Rudin's RCA tells you how to use polar coordinates in $mathbb R^{n}$.
– Kavi Rama Murthy
Nov 21 '18 at 7:48