Is the supremum of $int_{B(y,r)}lvert urvert^{p^ast}$ where $uin W_0^{1,p}$ attained inside an open set?
$begingroup$
Let $Omegasubseteq mathbb{R}^n$ be a bounded domain with smooth boundary. Fix $1< p < n$ and let $p^ast$ denote the sobolev conjugate of $p$.
Now, fix a small number $r>0$ and suppose $uin W_0^{1,p}(Omega)$. By the Sobolev inequality, we also know that $uin L^{p^ast}(Omega)$. Furthermore, extending $u$ to be $0$ outside of $Omega$ I may suppose that $uin W_0^{1,p}(mathbb{R}^n)cap L^{p^ast}(mathbb{R}^n)$.
Consider now
$$
sup_{yin mathbb{R}^n}int_{B(y,r)}lvert urvert^{p^ast}
$$
which we know to be finite since $uin L^{p^ast}(mathbb{R}^n)$. First, I was able to show that the supremum is achieved at some fixed point $yinmathbb{R}^n$. I can show this by taking a sequence $(y_m)$, considering a converging sub-sequence and then applying the dominated convergence theorem.
Is it possible to show that the supremum is acheived at some point $yin Omega$?
It seems intuitive that this should be the case but I am unsure how to prove it.
real-analysis pde sobolev-spaces lp-spaces
asked Dec 30 '18 at 20:02
QuokaQuoka
1,269312
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add a comment |
$begingroup$
Let $Omegasubseteq mathbb{R}^n$ be a bounded domain with smooth boundary. Fix $1< p < n$ and let $p^ast$ denote the sobolev conjugate of $p$.
Now, fix a small number $r>0$ and suppose $uin W_0^{1,p}(Omega)$. By the Sobolev inequality, we also know that $uin L^{p^ast}(Omega)$. Furthermore, extending $u$ to be $0$ outside of $Omega$ I may suppose that $uin W_0^{1,p}(mathbb{R}^n)cap L^{p^ast}(mathbb{R}^n)$.
Consider now
$$
sup_{yin mathbb{R}^n}int_{B(y,r)}lvert urvert^{p^ast}
$$
which we know to be finite since $uin L^{p^ast}(mathbb{R}^n)$. First, I was able to show that the supremum is achieved at some fixed point $yinmathbb{R}^n$. I can show this by taking a sequence $(y_m)$, considering a converging sub-sequence and then applying the dominated convergence theorem.
Is it possible to show that the supremum is acheived at some point $yin Omega$?
It seems intuitive that this should be the case but I am unsure how to prove it.
real-analysis pde sobolev-spaces lp-spaces
asked Dec 30 '18 at 20:02
QuokaQuoka
1,269312
$endgroup$
add a comment |
$begingroup$
Let $Omegasubseteq mathbb{R}^n$ be a bounded domain with smooth boundary. Fix $1< p < n$ and let $p^ast$ denote the sobolev conjugate of $p$.
Now, fix a small number $r>0$ and suppose $uin W_0^{1,p}(Omega)$. By the Sobolev inequality, we also know that $uin L^{p^ast}(Omega)$. Furthermore, extending $u$ to be $0$ outside of $Omega$ I may suppose that $uin W_0^{1,p}(mathbb{R}^n)cap L^{p^ast}(mathbb{R}^n)$.
Consider now
$$
sup_{yin mathbb{R}^n}int_{B(y,r)}lvert urvert^{p^ast}
$$
which we know to be finite since $uin L^{p^ast}(mathbb{R}^n)$. First, I was able to show that the supremum is achieved at some fixed point $yinmathbb{R}^n$. I can show this by taking a sequence $(y_m)$, considering a converging sub-sequence and then applying the dominated convergence theorem.
Is it possible to show that the supremum is acheived at some point $yin Omega$?
It seems intuitive that this should be the case but I am unsure how to prove it.
real-analysis pde sobolev-spaces lp-spaces
asked Dec 30 '18 at 20:02
QuokaQuoka
1,269312
$endgroup$
Let $Omegasubseteq mathbb{R}^n$ be a bounded domain with smooth boundary. Fix $1< p < n$ and let $p^ast$ denote the sobolev conjugate of $p$.
Now, fix a small number $r>0$ and suppose $uin W_0^{1,p}(Omega)$. By the Sobolev inequality, we also know that $uin L^{p^ast}(Omega)$. Furthermore, extending $u$ to be $0$ outside of $Omega$ I may suppose that $uin W_0^{1,p}(mathbb{R}^n)cap L^{p^ast}(mathbb{R}^n)$.
Consider now
$$
sup_{yin mathbb{R}^n}int_{B(y,r)}lvert urvert^{p^ast}
$$
which we know to be finite since $uin L^{p^ast}(mathbb{R}^n)$. First, I was able to show that the supremum is achieved at some fixed point $yinmathbb{R}^n$. I can show this by taking a sequence $(y_m)$, considering a converging sub-sequence and then applying the dominated convergence theorem.
Is it possible to show that the supremum is acheived at some point $yin Omega$?
It seems intuitive that this should be the case but I am unsure how to prove it.
real-analysis pde sobolev-spaces lp-spaces
real-analysis pde sobolev-spaces lp-spaces
asked Dec 30 '18 at 20:02
QuokaQuoka
1,269312
asked Dec 30 '18 at 20:02
QuokaQuoka
1,269312
edited Dec 30 '18 at 21:32
Quoka
asked Dec 30 '18 at 20:02
QuokaQuoka
1,269312
asked Dec 30 '18 at 20:02
QuokaQuoka
1,269312
asked Dec 30 '18 at 20:02
QuokaQuoka
1,269312
1,269312
add a comment |
add a comment |
1 Answer
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For $r$ fixed, here is a counterexample. Consider any $u$ with support $Omega = { 1le |x|le 2}$ and set $r=2$. Then the maximiser is at $x=0notinOmega$, as any deviation means you miss part of the support. It would seem then that you need to tune $r$ based on $Omega$ for a positive result.
answered Dec 30 '18 at 22:01
Calvin KhorCalvin Khor
12.1k21438
$endgroup$
$begingroup$
That's surprisingly simple! Thank you :-)
$endgroup$
– Quoka
Dec 30 '18 at 22:14
$begingroup$
@Quoka you're welcome :)
$endgroup$
– Calvin Khor
Dec 30 '18 at 22:21
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
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active
oldest
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active
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$begingroup$
For $r$ fixed, here is a counterexample. Consider any $u$ with support $Omega = { 1le |x|le 2}$ and set $r=2$. Then the maximiser is at $x=0notinOmega$, as any deviation means you miss part of the support. It would seem then that you need to tune $r$ based on $Omega$ for a positive result.
answered Dec 30 '18 at 22:01
Calvin KhorCalvin Khor
12.1k21438
$endgroup$
$begingroup$
That's surprisingly simple! Thank you :-)
$endgroup$
– Quoka
Dec 30 '18 at 22:14
$begingroup$
@Quoka you're welcome :)
$endgroup$
– Calvin Khor
Dec 30 '18 at 22:21
add a comment |
$begingroup$
For $r$ fixed, here is a counterexample. Consider any $u$ with support $Omega = { 1le |x|le 2}$ and set $r=2$. Then the maximiser is at $x=0notinOmega$, as any deviation means you miss part of the support. It would seem then that you need to tune $r$ based on $Omega$ for a positive result.
answered Dec 30 '18 at 22:01
Calvin KhorCalvin Khor
12.1k21438
$endgroup$
$begingroup$
That's surprisingly simple! Thank you :-)
$endgroup$
– Quoka
Dec 30 '18 at 22:14
$begingroup$
@Quoka you're welcome :)
$endgroup$
– Calvin Khor
Dec 30 '18 at 22:21
add a comment |
$begingroup$
For $r$ fixed, here is a counterexample. Consider any $u$ with support $Omega = { 1le |x|le 2}$ and set $r=2$. Then the maximiser is at $x=0notinOmega$, as any deviation means you miss part of the support. It would seem then that you need to tune $r$ based on $Omega$ for a positive result.
answered Dec 30 '18 at 22:01
Calvin KhorCalvin Khor
12.1k21438
$endgroup$
For $r$ fixed, here is a counterexample. Consider any $u$ with support $Omega = { 1le |x|le 2}$ and set $r=2$. Then the maximiser is at $x=0notinOmega$, as any deviation means you miss part of the support. It would seem then that you need to tune $r$ based on $Omega$ for a positive result.
answered Dec 30 '18 at 22:01
Calvin KhorCalvin Khor
12.1k21438
answered Dec 30 '18 at 22:01
Calvin KhorCalvin Khor
12.1k21438
answered Dec 30 '18 at 22:01
Calvin KhorCalvin Khor
12.1k21438
answered Dec 30 '18 at 22:01
Calvin KhorCalvin Khor
12.1k21438
12.1k21438
$begingroup$
That's surprisingly simple! Thank you :-)
$endgroup$
– Quoka
Dec 30 '18 at 22:14
$begingroup$
@Quoka you're welcome :)
$endgroup$
– Calvin Khor
Dec 30 '18 at 22:21
add a comment |
$begingroup$
That's surprisingly simple! Thank you :-)
$endgroup$
– Quoka
Dec 30 '18 at 22:14
$begingroup$
@Quoka you're welcome :)
$endgroup$
– Calvin Khor
Dec 30 '18 at 22:21
$begingroup$
That's surprisingly simple! Thank you :-)
$endgroup$
– Quoka
Dec 30 '18 at 22:14
$begingroup$
That's surprisingly simple! Thank you :-)
$endgroup$
– Quoka
Dec 30 '18 at 22:14
$begingroup$
@Quoka you're welcome :)
$endgroup$
– Calvin Khor
Dec 30 '18 at 22:21
$begingroup$
@Quoka you're welcome :)
$endgroup$
– Calvin Khor
Dec 30 '18 at 22:21
add a comment |
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