Integral in termsof supremum











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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







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  • 1




    If $w=0$ this cannot be true.
    – daw
    12 hours ago






  • 1




    For $A_p$ weights $w$ of course $w$ is nonzero (a.e.), so the same proof works.
    – user10354138
    11 hours ago










  • Thank you very much.
    – Mathlover
    7 hours ago










  • Can you kindly have a look at the foolwing question in the link : math.stackexchange.com/questions/2996709/…
    – Mathlover
    6 hours ago















up vote
0
down vote

favorite
1












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







share|cite|improve this question




















  • 1




    If $w=0$ this cannot be true.
    – daw
    12 hours ago






  • 1




    For $A_p$ weights $w$ of course $w$ is nonzero (a.e.), so the same proof works.
    – user10354138
    11 hours ago










  • Thank you very much.
    – Mathlover
    7 hours ago










  • Can you kindly have a look at the foolwing question in the link : math.stackexchange.com/questions/2996709/…
    – Mathlover
    6 hours ago













up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited 11 hours ago









user10354138

6,124523




6,124523










asked 13 hours ago









Mathlover

937




937








  • 1




    If $w=0$ this cannot be true.
    – daw
    12 hours ago






  • 1




    For $A_p$ weights $w$ of course $w$ is nonzero (a.e.), so the same proof works.
    – user10354138
    11 hours ago










  • Thank you very much.
    – Mathlover
    7 hours ago










  • Can you kindly have a look at the foolwing question in the link : math.stackexchange.com/questions/2996709/…
    – Mathlover
    6 hours ago














  • 1




    If $w=0$ this cannot be true.
    – daw
    12 hours ago






  • 1




    For $A_p$ weights $w$ of course $w$ is nonzero (a.e.), so the same proof works.
    – user10354138
    11 hours ago










  • Thank you very much.
    – Mathlover
    7 hours ago










  • Can you kindly have a look at the foolwing question in the link : math.stackexchange.com/questions/2996709/…
    – Mathlover
    6 hours ago








1




1




If $w=0$ this cannot be true.
– daw
12 hours ago




If $w=0$ this cannot be true.
– daw
12 hours ago




1




1




For $A_p$ weights $w$ of course $w$ is nonzero (a.e.), so the same proof works.
– user10354138
11 hours ago




For $A_p$ weights $w$ of course $w$ is nonzero (a.e.), so the same proof works.
– user10354138
11 hours ago












Thank you very much.
– Mathlover
7 hours ago




Thank you very much.
– Mathlover
7 hours ago












Can you kindly have a look at the foolwing question in the link : math.stackexchange.com/questions/2996709/…
– Mathlover
6 hours ago




Can you kindly have a look at the foolwing question in the link : math.stackexchange.com/questions/2996709/…
– Mathlover
6 hours ago















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