Is $|x|^{-alpha}$ integrable for polynomially bounded measures on $mathbb{R}^n$
We know that $|x|^{-alpha}$ is in $L^1 (xin mathbb{R}^n:|x| ge 1)$ with the normal Lebesgue measure for $alpha > n$. But what if we had a measure $mu$ on $mathbb{R}^n$ which is polynomially bounded, i.e., $mu(|x|le A) le C(1+A^N)$ where $C,N$ are fixed constants, then would we have something like $|x|^{-alpha}$ is in $L^1 ({xin mathbb{R}^n:|x| ge 1},mu)$ for $alpha >N$?
measure-theory lebesgue-measure
asked Nov 30 '18 at 1:10
Andrew YuanAndrew Yuan
1508
add a comment |
We know that $|x|^{-alpha}$ is in $L^1 (xin mathbb{R}^n:|x| ge 1)$ with the normal Lebesgue measure for $alpha > n$. But what if we had a measure $mu$ on $mathbb{R}^n$ which is polynomially bounded, i.e., $mu(|x|le A) le C(1+A^N)$ where $C,N$ are fixed constants, then would we have something like $|x|^{-alpha}$ is in $L^1 ({xin mathbb{R}^n:|x| ge 1},mu)$ for $alpha >N$?
measure-theory lebesgue-measure
asked Nov 30 '18 at 1:10
Andrew YuanAndrew Yuan
1508
add a comment |
We know that $|x|^{-alpha}$ is in $L^1 (xin mathbb{R}^n:|x| ge 1)$ with the normal Lebesgue measure for $alpha > n$. But what if we had a measure $mu$ on $mathbb{R}^n$ which is polynomially bounded, i.e., $mu(|x|le A) le C(1+A^N)$ where $C,N$ are fixed constants, then would we have something like $|x|^{-alpha}$ is in $L^1 ({xin mathbb{R}^n:|x| ge 1},mu)$ for $alpha >N$?
measure-theory lebesgue-measure
asked Nov 30 '18 at 1:10
Andrew YuanAndrew Yuan
1508
We know that $|x|^{-alpha}$ is in $L^1 (xin mathbb{R}^n:|x| ge 1)$ with the normal Lebesgue measure for $alpha > n$. But what if we had a measure $mu$ on $mathbb{R}^n$ which is polynomially bounded, i.e., $mu(|x|le A) le C(1+A^N)$ where $C,N$ are fixed constants, then would we have something like $|x|^{-alpha}$ is in $L^1 ({xin mathbb{R}^n:|x| ge 1},mu)$ for $alpha >N$?
measure-theory lebesgue-measure
measure-theory lebesgue-measure
asked Nov 30 '18 at 1:10
Andrew YuanAndrew Yuan
1508
asked Nov 30 '18 at 1:10
Andrew YuanAndrew Yuan
1508
asked Nov 30 '18 at 1:10
Andrew YuanAndrew Yuan
1508
asked Nov 30 '18 at 1:10
Andrew YuanAndrew Yuan
1508
asked Nov 30 '18 at 1:10
Andrew YuanAndrew Yuan
1508
1508
add a comment |
add a comment |
1 Answer
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We split the integral into dyadic pieces in the following way:
begin{align}
int_{{|x| ge 1}} |x|^{-alpha} , mathrm{d}mu(x) &= sum_{n=0}^infty int_{{2^{n+1} > |x| ge 2^n}} |x|^{-alpha} , mathrm{d}mu(x) \
&le sum_{n=0}^infty 2^{-nalpha} mu{2^n le |x| < 2^{n+1}}
end{align}
Now we can use the polynomially growth bound $mu(|x| le 2^{n+1}) le C(1+2^{(n+1)N})$ to get that the last term is bounded by
begin{align}
sum_{n=0}^infty 2^{-nalpha} mu{|x| le 2^{n+1}} le C sum_{n=0}^infty 2^{-nalpha} (1+ 2^{(n+1)N}).
end{align}
Here the last sum is convergent if and only if $alpha >N$ and $alpha >0$. Thus, in this more general case, we have also integrability provided that $alpha > N,$ where I supposed that $N >0$.
answered Nov 30 '18 at 9:22
p4schp4sch
4,770217
Notice you can bypass the dyadic partitioning by setting up the first sum over dyadic ranges.
– Guacho Perez
Nov 30 '18 at 15:47
I have simplified the proof with regard to your comment. Now the proof is simpler.
– p4sch
Nov 30 '18 at 22:09
nice proof, (+1).
– Guacho Perez
Nov 30 '18 at 22:11
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
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active
oldest
votes
We split the integral into dyadic pieces in the following way:
begin{align}
int_{{|x| ge 1}} |x|^{-alpha} , mathrm{d}mu(x) &= sum_{n=0}^infty int_{{2^{n+1} > |x| ge 2^n}} |x|^{-alpha} , mathrm{d}mu(x) \
&le sum_{n=0}^infty 2^{-nalpha} mu{2^n le |x| < 2^{n+1}}
end{align}
Now we can use the polynomially growth bound $mu(|x| le 2^{n+1}) le C(1+2^{(n+1)N})$ to get that the last term is bounded by
begin{align}
sum_{n=0}^infty 2^{-nalpha} mu{|x| le 2^{n+1}} le C sum_{n=0}^infty 2^{-nalpha} (1+ 2^{(n+1)N}).
end{align}
Here the last sum is convergent if and only if $alpha >N$ and $alpha >0$. Thus, in this more general case, we have also integrability provided that $alpha > N,$ where I supposed that $N >0$.
answered Nov 30 '18 at 9:22
p4schp4sch
4,770217
Notice you can bypass the dyadic partitioning by setting up the first sum over dyadic ranges.
– Guacho Perez
Nov 30 '18 at 15:47
I have simplified the proof with regard to your comment. Now the proof is simpler.
– p4sch
Nov 30 '18 at 22:09
nice proof, (+1).
– Guacho Perez
Nov 30 '18 at 22:11
add a comment |
We split the integral into dyadic pieces in the following way:
begin{align}
int_{{|x| ge 1}} |x|^{-alpha} , mathrm{d}mu(x) &= sum_{n=0}^infty int_{{2^{n+1} > |x| ge 2^n}} |x|^{-alpha} , mathrm{d}mu(x) \
&le sum_{n=0}^infty 2^{-nalpha} mu{2^n le |x| < 2^{n+1}}
end{align}
Now we can use the polynomially growth bound $mu(|x| le 2^{n+1}) le C(1+2^{(n+1)N})$ to get that the last term is bounded by
begin{align}
sum_{n=0}^infty 2^{-nalpha} mu{|x| le 2^{n+1}} le C sum_{n=0}^infty 2^{-nalpha} (1+ 2^{(n+1)N}).
end{align}
Here the last sum is convergent if and only if $alpha >N$ and $alpha >0$. Thus, in this more general case, we have also integrability provided that $alpha > N,$ where I supposed that $N >0$.
answered Nov 30 '18 at 9:22
p4schp4sch
4,770217
Notice you can bypass the dyadic partitioning by setting up the first sum over dyadic ranges.
– Guacho Perez
Nov 30 '18 at 15:47
I have simplified the proof with regard to your comment. Now the proof is simpler.
– p4sch
Nov 30 '18 at 22:09
nice proof, (+1).
– Guacho Perez
Nov 30 '18 at 22:11
add a comment |
We split the integral into dyadic pieces in the following way:
begin{align}
int_{{|x| ge 1}} |x|^{-alpha} , mathrm{d}mu(x) &= sum_{n=0}^infty int_{{2^{n+1} > |x| ge 2^n}} |x|^{-alpha} , mathrm{d}mu(x) \
&le sum_{n=0}^infty 2^{-nalpha} mu{2^n le |x| < 2^{n+1}}
end{align}
Now we can use the polynomially growth bound $mu(|x| le 2^{n+1}) le C(1+2^{(n+1)N})$ to get that the last term is bounded by
begin{align}
sum_{n=0}^infty 2^{-nalpha} mu{|x| le 2^{n+1}} le C sum_{n=0}^infty 2^{-nalpha} (1+ 2^{(n+1)N}).
end{align}
Here the last sum is convergent if and only if $alpha >N$ and $alpha >0$. Thus, in this more general case, we have also integrability provided that $alpha > N,$ where I supposed that $N >0$.
answered Nov 30 '18 at 9:22
p4schp4sch
4,770217
We split the integral into dyadic pieces in the following way:
begin{align}
int_{{|x| ge 1}} |x|^{-alpha} , mathrm{d}mu(x) &= sum_{n=0}^infty int_{{2^{n+1} > |x| ge 2^n}} |x|^{-alpha} , mathrm{d}mu(x) \
&le sum_{n=0}^infty 2^{-nalpha} mu{2^n le |x| < 2^{n+1}}
end{align}
Now we can use the polynomially growth bound $mu(|x| le 2^{n+1}) le C(1+2^{(n+1)N})$ to get that the last term is bounded by
begin{align}
sum_{n=0}^infty 2^{-nalpha} mu{|x| le 2^{n+1}} le C sum_{n=0}^infty 2^{-nalpha} (1+ 2^{(n+1)N}).
end{align}
Here the last sum is convergent if and only if $alpha >N$ and $alpha >0$. Thus, in this more general case, we have also integrability provided that $alpha > N,$ where I supposed that $N >0$.
answered Nov 30 '18 at 9:22
p4schp4sch
4,770217
edited Nov 30 '18 at 22:08
answered Nov 30 '18 at 9:22
p4schp4sch
4,770217
answered Nov 30 '18 at 9:22
p4schp4sch
4,770217
answered Nov 30 '18 at 9:22
p4schp4sch
4,770217
4,770217
Notice you can bypass the dyadic partitioning by setting up the first sum over dyadic ranges.
– Guacho Perez
Nov 30 '18 at 15:47
I have simplified the proof with regard to your comment. Now the proof is simpler.
– p4sch
Nov 30 '18 at 22:09
nice proof, (+1).
– Guacho Perez
Nov 30 '18 at 22:11
add a comment |
Notice you can bypass the dyadic partitioning by setting up the first sum over dyadic ranges.
– Guacho Perez
Nov 30 '18 at 15:47
I have simplified the proof with regard to your comment. Now the proof is simpler.
– p4sch
Nov 30 '18 at 22:09
nice proof, (+1).
– Guacho Perez
Nov 30 '18 at 22:11
Notice you can bypass the dyadic partitioning by setting up the first sum over dyadic ranges.
– Guacho Perez
Nov 30 '18 at 15:47
Notice you can bypass the dyadic partitioning by setting up the first sum over dyadic ranges.
– Guacho Perez
Nov 30 '18 at 15:47
I have simplified the proof with regard to your comment. Now the proof is simpler.
– p4sch
Nov 30 '18 at 22:09
I have simplified the proof with regard to your comment. Now the proof is simpler.
– p4sch
Nov 30 '18 at 22:09
nice proof, (+1).
– Guacho Perez
Nov 30 '18 at 22:11
nice proof, (+1).
– Guacho Perez
Nov 30 '18 at 22:11
add a comment |
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