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How to extend results from space of smooth functions to Sobolev spaces?
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I regularly see proofs involving Sobolev spaces where the proof states it will show some result holds for, say, $u in H_0^1(Omega)$ where $Omega$ is a smooth bounded domain.
Then right away it will say that it suffices to the result holds for $u in C^infty(Omega)$. So we have proved that the result holds for $u in C^infty(Omega)$, how then can we show rigorously that this result also holds for $u in H_0^1(Omega)$?
functional-analysis pde sobolev-spaces
asked Jul 19 at 11:19
sonicboom
3,44182550
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up vote
3
down vote
favorite
I regularly see proofs involving Sobolev spaces where the proof states it will show some result holds for, say, $u in H_0^1(Omega)$ where $Omega$ is a smooth bounded domain.
Then right away it will say that it suffices to the result holds for $u in C^infty(Omega)$. So we have proved that the result holds for $u in C^infty(Omega)$, how then can we show rigorously that this result also holds for $u in H_0^1(Omega)$?
functional-analysis pde sobolev-spaces
asked Jul 19 at 11:19
sonicboom
3,44182550
2
In most case they use the fact that $C_0^infty(Omega)$ is dense in $H^1_0(Omega)$.
– John Ma
Jul 19 at 11:34
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I regularly see proofs involving Sobolev spaces where the proof states it will show some result holds for, say, $u in H_0^1(Omega)$ where $Omega$ is a smooth bounded domain.
Then right away it will say that it suffices to the result holds for $u in C^infty(Omega)$. So we have proved that the result holds for $u in C^infty(Omega)$, how then can we show rigorously that this result also holds for $u in H_0^1(Omega)$?
functional-analysis pde sobolev-spaces
asked Jul 19 at 11:19
sonicboom
3,44182550
I regularly see proofs involving Sobolev spaces where the proof states it will show some result holds for, say, $u in H_0^1(Omega)$ where $Omega$ is a smooth bounded domain.
Then right away it will say that it suffices to the result holds for $u in C^infty(Omega)$. So we have proved that the result holds for $u in C^infty(Omega)$, how then can we show rigorously that this result also holds for $u in H_0^1(Omega)$?
functional-analysis pde sobolev-spaces
asked Jul 19 at 11:19
sonicboom
3,44182550
asked Jul 19 at 11:19
sonicboom
3,44182550
asked Jul 19 at 11:19
sonicboom
3,44182550
asked Jul 19 at 11:19
sonicboom
3,44182550
3,44182550
2
In most case they use the fact that $C_0^infty(Omega)$ is dense in $H^1_0(Omega)$.
– John Ma
Jul 19 at 11:34
add a comment |Â
2
In most case they use the fact that $C_0^infty(Omega)$ is dense in $H^1_0(Omega)$.
– John Ma
Jul 19 at 11:34
2
2
In most case they use the fact that $C_0^infty(Omega)$ is dense in $H^1_0(Omega)$.
– John Ma
Jul 19 at 11:34
In most case they use the fact that $C_0^infty(Omega)$ is dense in $H^1_0(Omega)$.
– John Ma
Jul 19 at 11:34
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
5
down vote
accepted
The result usually has the form $F(u) = G(u)$ where $F$ and $G$ are some functionals on the Sobolev space. If we establish that
- $F$ and $G$ are continuous with respect to the Sobolev norm
- $F(u) = G(u)$ for all $u$ in some dense subset of the Sobolev space
then the conclusion $Fequiv G$ follows in the usual way: $f,g$ continuous from $X$ to $Y$. if they are agree on a dense set $A$ of $X$ then they agree on $X$
The set of smooth compactly supported functions is dense in $H^1_0$. The continuity of $F$ and $G$ is often glossed over in proofs, as "easy to see" because $F$ and $G$ are made of things that are controlled by the Sobolev norm.
answered Jul 19 at 17:07
user357151
13.9k31140
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
The result usually has the form $F(u) = G(u)$ where $F$ and $G$ are some functionals on the Sobolev space. If we establish that
- $F$ and $G$ are continuous with respect to the Sobolev norm
- $F(u) = G(u)$ for all $u$ in some dense subset of the Sobolev space
then the conclusion $Fequiv G$ follows in the usual way: $f,g$ continuous from $X$ to $Y$. if they are agree on a dense set $A$ of $X$ then they agree on $X$
The set of smooth compactly supported functions is dense in $H^1_0$. The continuity of $F$ and $G$ is often glossed over in proofs, as "easy to see" because $F$ and $G$ are made of things that are controlled by the Sobolev norm.
answered Jul 19 at 17:07
user357151
13.9k31140
add a comment |Â
up vote
5
down vote
accepted
The result usually has the form $F(u) = G(u)$ where $F$ and $G$ are some functionals on the Sobolev space. If we establish that
- $F$ and $G$ are continuous with respect to the Sobolev norm
- $F(u) = G(u)$ for all $u$ in some dense subset of the Sobolev space
then the conclusion $Fequiv G$ follows in the usual way: $f,g$ continuous from $X$ to $Y$. if they are agree on a dense set $A$ of $X$ then they agree on $X$
The set of smooth compactly supported functions is dense in $H^1_0$. The continuity of $F$ and $G$ is often glossed over in proofs, as "easy to see" because $F$ and $G$ are made of things that are controlled by the Sobolev norm.
answered Jul 19 at 17:07
user357151
13.9k31140
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
The result usually has the form $F(u) = G(u)$ where $F$ and $G$ are some functionals on the Sobolev space. If we establish that
- $F$ and $G$ are continuous with respect to the Sobolev norm
- $F(u) = G(u)$ for all $u$ in some dense subset of the Sobolev space
then the conclusion $Fequiv G$ follows in the usual way: $f,g$ continuous from $X$ to $Y$. if they are agree on a dense set $A$ of $X$ then they agree on $X$
The set of smooth compactly supported functions is dense in $H^1_0$. The continuity of $F$ and $G$ is often glossed over in proofs, as "easy to see" because $F$ and $G$ are made of things that are controlled by the Sobolev norm.
answered Jul 19 at 17:07
user357151
13.9k31140
The result usually has the form $F(u) = G(u)$ where $F$ and $G$ are some functionals on the Sobolev space. If we establish that
- $F$ and $G$ are continuous with respect to the Sobolev norm
- $F(u) = G(u)$ for all $u$ in some dense subset of the Sobolev space
then the conclusion $Fequiv G$ follows in the usual way: $f,g$ continuous from $X$ to $Y$. if they are agree on a dense set $A$ of $X$ then they agree on $X$
The set of smooth compactly supported functions is dense in $H^1_0$. The continuity of $F$ and $G$ is often glossed over in proofs, as "easy to see" because $F$ and $G$ are made of things that are controlled by the Sobolev norm.
answered Jul 19 at 17:07
user357151
13.9k31140
answered Jul 19 at 17:07
user357151
13.9k31140
answered Jul 19 at 17:07
user357151
13.9k31140
answered Jul 19 at 17:07
user357151
13.9k31140
13.9k31140
add a comment |Â
add a comment |Â
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2
In most case they use the fact that $C_0^infty(Omega)$ is dense in $H^1_0(Omega)$.
– John Ma
Jul 19 at 11:34