Integral in termsof supremum
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1
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I am facing difficulty to prove the following fact:
If $w$ is a locally integrable positive function in $Omega$, then
$$
sup_{B(x,R)}lvert vrvert=lim_{ptoinfty}lVert vrVert_{L^p(B(x,R),w))},
$$
where $B(x,R)subset Omega$ and $Omega$ is a bounded domain, and $lVert vrVert_{L^p(B(x,R),w))}$ is defined to be
$$
left(int_{B(x,R)}lvert vrvert^pw(x),dxright)^{1/p}.
$$
For $w=1$, I know this is true. But for non-constant $w$, even for $A_p$ weights does the same hold true? I have sen this fact is applied in the paper attached in the link: see page 15 or 16.
http://imar.ro/journals/Mathematical_Reports/Pdfs/2017/3/3.pdf
real-analysis functional-analysis pde sobolev-spaces
edited Nov 13 at 8:20
user10354138
6,294623
asked Nov 13 at 6:52
Mathlover
1038
add a comment |
up vote
1
down vote
favorite
I am facing difficulty to prove the following fact:
If $w$ is a locally integrable positive function in $Omega$, then
$$
sup_{B(x,R)}lvert vrvert=lim_{ptoinfty}lVert vrVert_{L^p(B(x,R),w))},
$$
where $B(x,R)subset Omega$ and $Omega$ is a bounded domain, and $lVert vrVert_{L^p(B(x,R),w))}$ is defined to be
$$
left(int_{B(x,R)}lvert vrvert^pw(x),dxright)^{1/p}.
$$
For $w=1$, I know this is true. But for non-constant $w$, even for $A_p$ weights does the same hold true? I have sen this fact is applied in the paper attached in the link: see page 15 or 16.
http://imar.ro/journals/Mathematical_Reports/Pdfs/2017/3/3.pdf
real-analysis functional-analysis pde sobolev-spaces
edited Nov 13 at 8:20
user10354138
6,294623
asked Nov 13 at 6:52
Mathlover
1038
1
If $w=0$ this cannot be true.
– daw
Nov 13 at 7:48
1
For $A_p$ weights $w$ of course $w$ is nonzero (a.e.), so the same proof works.
– user10354138
Nov 13 at 8:53
Thank you very much.
– Mathlover
Nov 13 at 12:31
Can you kindly have a look at the foolwing question in the link : math.stackexchange.com/questions/2996709/…
– Mathlover
Nov 13 at 13:07
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am facing difficulty to prove the following fact:
If $w$ is a locally integrable positive function in $Omega$, then
$$
sup_{B(x,R)}lvert vrvert=lim_{ptoinfty}lVert vrVert_{L^p(B(x,R),w))},
$$
where $B(x,R)subset Omega$ and $Omega$ is a bounded domain, and $lVert vrVert_{L^p(B(x,R),w))}$ is defined to be
$$
left(int_{B(x,R)}lvert vrvert^pw(x),dxright)^{1/p}.
$$
For $w=1$, I know this is true. But for non-constant $w$, even for $A_p$ weights does the same hold true? I have sen this fact is applied in the paper attached in the link: see page 15 or 16.
http://imar.ro/journals/Mathematical_Reports/Pdfs/2017/3/3.pdf
real-analysis functional-analysis pde sobolev-spaces
edited Nov 13 at 8:20
user10354138
6,294623
asked Nov 13 at 6:52
Mathlover
1038
I am facing difficulty to prove the following fact:
If $w$ is a locally integrable positive function in $Omega$, then
$$
sup_{B(x,R)}lvert vrvert=lim_{ptoinfty}lVert vrVert_{L^p(B(x,R),w))},
$$
where $B(x,R)subset Omega$ and $Omega$ is a bounded domain, and $lVert vrVert_{L^p(B(x,R),w))}$ is defined to be
$$
left(int_{B(x,R)}lvert vrvert^pw(x),dxright)^{1/p}.
$$
For $w=1$, I know this is true. But for non-constant $w$, even for $A_p$ weights does the same hold true? I have sen this fact is applied in the paper attached in the link: see page 15 or 16.
http://imar.ro/journals/Mathematical_Reports/Pdfs/2017/3/3.pdf
real-analysis functional-analysis pde sobolev-spaces
real-analysis functional-analysis pde sobolev-spaces
edited Nov 13 at 8:20
user10354138
6,294623
asked Nov 13 at 6:52
Mathlover
1038
edited Nov 13 at 8:20
user10354138
6,294623
asked Nov 13 at 6:52
Mathlover
1038
edited Nov 13 at 8:20
user10354138
6,294623
edited Nov 13 at 8:20
user10354138
6,294623
edited Nov 13 at 8:20
user10354138
6,294623
6,294623
asked Nov 13 at 6:52
Mathlover
1038
asked Nov 13 at 6:52
Mathlover
1038
asked Nov 13 at 6:52
Mathlover
1038
1038
1
If $w=0$ this cannot be true.
– daw
Nov 13 at 7:48
1
For $A_p$ weights $w$ of course $w$ is nonzero (a.e.), so the same proof works.
– user10354138
Nov 13 at 8:53
Thank you very much.
– Mathlover
Nov 13 at 12:31
Can you kindly have a look at the foolwing question in the link : math.stackexchange.com/questions/2996709/…
– Mathlover
Nov 13 at 13:07
add a comment |
1
If $w=0$ this cannot be true.
– daw
Nov 13 at 7:48
1
For $A_p$ weights $w$ of course $w$ is nonzero (a.e.), so the same proof works.
– user10354138
Nov 13 at 8:53
Thank you very much.
– Mathlover
Nov 13 at 12:31
Can you kindly have a look at the foolwing question in the link : math.stackexchange.com/questions/2996709/…
– Mathlover
Nov 13 at 13:07
1
1
If $w=0$ this cannot be true.
– daw
Nov 13 at 7:48
If $w=0$ this cannot be true.
– daw
Nov 13 at 7:48
1
1
For $A_p$ weights $w$ of course $w$ is nonzero (a.e.), so the same proof works.
– user10354138
Nov 13 at 8:53
For $A_p$ weights $w$ of course $w$ is nonzero (a.e.), so the same proof works.
– user10354138
Nov 13 at 8:53
Thank you very much.
– Mathlover
Nov 13 at 12:31
Thank you very much.
– Mathlover
Nov 13 at 12:31
Can you kindly have a look at the foolwing question in the link : math.stackexchange.com/questions/2996709/…
– Mathlover
Nov 13 at 13:07
Can you kindly have a look at the foolwing question in the link : math.stackexchange.com/questions/2996709/…
– Mathlover
Nov 13 at 13:07
add a comment |
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1
If $w=0$ this cannot be true.
– daw
Nov 13 at 7:48
1
For $A_p$ weights $w$ of course $w$ is nonzero (a.e.), so the same proof works.
– user10354138
Nov 13 at 8:53
Thank you very much.
– Mathlover
Nov 13 at 12:31
Can you kindly have a look at the foolwing question in the link : math.stackexchange.com/questions/2996709/…
– Mathlover
Nov 13 at 13:07