$H^1(Omega')$ 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.
functional-analysis pde sobolev-spaces
edited Nov 22 at 12:27
gerw
18.9k11133
asked Nov 22 at 12:24
gmirsan
63
add a comment |
up vote
0
down vote
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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.
functional-analysis pde sobolev-spaces
edited Nov 22 at 12:27
gerw
18.9k11133
asked Nov 22 at 12:24
gmirsan
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
add a comment |
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.
functional-analysis pde sobolev-spaces
edited Nov 22 at 12:27
gerw
18.9k11133
asked Nov 22 at 12:24
gmirsan
63
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
functional-analysis pde sobolev-spaces
edited Nov 22 at 12:27
gerw
18.9k11133
asked Nov 22 at 12:24
gmirsan
63
edited Nov 22 at 12:27
gerw
18.9k11133
asked Nov 22 at 12:24
gmirsan
63
edited Nov 22 at 12:27
gerw
18.9k11133
edited Nov 22 at 12:27
gerw
18.9k11133
edited Nov 22 at 12:27
gerw
18.9k11133
18.9k11133
asked Nov 22 at 12:24
gmirsan
63
asked Nov 22 at 12:24
gmirsan
63
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
add a comment |
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
add a comment |
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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