$H^1(Omega')$ is in genral not a subspace of $H^1 (Omega)$ for bounded domains $Omega' subset Omega$











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Reading about Sobolev spaces I found the following statement:



$H^1 ({Omega}')$ is not a subspace of $H^1(Omega)$ for $Omega'subset Omega$.



$left(text{However } H^1_0 (Omega ')subset H^1_0(Omega)right)$



I guess that the reason behind is that you can not control $fin H^1(Omega ')$ outside $Omega'$. But I'm having a hard time thinking a counterexample or a more rigorous explanation.










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  • 1




    can't you take something that blows up at the boundary?
    – mathworker21
    Nov 22 at 12:44










  • First of all, a function on $Omega'$ is not a function on $Omega$, so one has to say something to make sense of the inclusion in the first place. This is also the case (although easy) for $H^1_0 (Omega ')subset H^1_0(Omega)$. Please say something.
    – Michał Miśkiewicz
    Nov 22 at 15:36

















up vote
0
down vote

favorite












Reading about Sobolev spaces I found the following statement:



$H^1 ({Omega}')$ is not a subspace of $H^1(Omega)$ for $Omega'subset Omega$.



$left(text{However } H^1_0 (Omega ')subset H^1_0(Omega)right)$



I guess that the reason behind is that you can not control $fin H^1(Omega ')$ outside $Omega'$. But I'm having a hard time thinking a counterexample or a more rigorous explanation.










share|cite|improve this question




















  • 1




    can't you take something that blows up at the boundary?
    – mathworker21
    Nov 22 at 12:44










  • First of all, a function on $Omega'$ is not a function on $Omega$, so one has to say something to make sense of the inclusion in the first place. This is also the case (although easy) for $H^1_0 (Omega ')subset H^1_0(Omega)$. Please say something.
    – Michał Miśkiewicz
    Nov 22 at 15:36















up vote
0
down vote

favorite









up vote
0
down vote

favorite











Reading about Sobolev spaces I found the following statement:



$H^1 ({Omega}')$ is not a subspace of $H^1(Omega)$ for $Omega'subset Omega$.



$left(text{However } H^1_0 (Omega ')subset H^1_0(Omega)right)$



I guess that the reason behind is that you can not control $fin H^1(Omega ')$ outside $Omega'$. But I'm having a hard time thinking a counterexample or a more rigorous explanation.










share|cite|improve this question















Reading about Sobolev spaces I found the following statement:



$H^1 ({Omega}')$ is not a subspace of $H^1(Omega)$ for $Omega'subset Omega$.



$left(text{However } H^1_0 (Omega ')subset H^1_0(Omega)right)$



I guess that the reason behind is that you can not control $fin H^1(Omega ')$ outside $Omega'$. But I'm having a hard time thinking a counterexample or a more rigorous explanation.







functional-analysis pde sobolev-spaces






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edited Nov 22 at 12:27









gerw

18.9k11133




18.9k11133










asked Nov 22 at 12:24









gmirsan

63




63








  • 1




    can't you take something that blows up at the boundary?
    – mathworker21
    Nov 22 at 12:44










  • First of all, a function on $Omega'$ is not a function on $Omega$, so one has to say something to make sense of the inclusion in the first place. This is also the case (although easy) for $H^1_0 (Omega ')subset H^1_0(Omega)$. Please say something.
    – Michał Miśkiewicz
    Nov 22 at 15:36
















  • 1




    can't you take something that blows up at the boundary?
    – mathworker21
    Nov 22 at 12:44










  • First of all, a function on $Omega'$ is not a function on $Omega$, so one has to say something to make sense of the inclusion in the first place. This is also the case (although easy) for $H^1_0 (Omega ')subset H^1_0(Omega)$. Please say something.
    – Michał Miśkiewicz
    Nov 22 at 15:36










1




1




can't you take something that blows up at the boundary?
– mathworker21
Nov 22 at 12:44




can't you take something that blows up at the boundary?
– mathworker21
Nov 22 at 12:44












First of all, a function on $Omega'$ is not a function on $Omega$, so one has to say something to make sense of the inclusion in the first place. This is also the case (although easy) for $H^1_0 (Omega ')subset H^1_0(Omega)$. Please say something.
– Michał Miśkiewicz
Nov 22 at 15:36






First of all, a function on $Omega'$ is not a function on $Omega$, so one has to say something to make sense of the inclusion in the first place. This is also the case (although easy) for $H^1_0 (Omega ')subset H^1_0(Omega)$. Please say something.
– Michał Miśkiewicz
Nov 22 at 15:36

















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