Extended Global Approximation Theorem
up vote
1
down vote
favorite
In Evans,
$textbf{Theorem} $ (Global Approximation Theorem) Assume $U$ is bounded, and $partial U$ is $C^1$. Suppose as well that $u in W^{k,p}(U)$ for some $1leq p < infty$. Then, there exist functions $u_m in C^{infty}(bar{U})$ such that
begin{align*}
u_m rightarrow u quad textrm{ in } W^{k,p}(U)
end{align*}
$textbf{Question}$ Although we change the boundary condition like
begin{align*}
partial U=bigcup_{j=1}^n Gamma_j, quad (textrm{boundary is piecewise } C^{1})
end{align*}
where each $Gamma_j$ for $j=1, cdots, n$ is a $C^1$, $Gamma_j$ and $Gamma_{j^{'}}$ do not intersect except at their endpoints if $jneq j'$, then does the theorem still hold?
Any help is appreciated!!
I want to know references related that...
Thank you!!
analysis pde sobolev-spaces approximation-theory
asked Nov 14 at 9:19
w.sdka
32119
add a comment |
up vote
1
down vote
favorite
In Evans,
$textbf{Theorem} $ (Global Approximation Theorem) Assume $U$ is bounded, and $partial U$ is $C^1$. Suppose as well that $u in W^{k,p}(U)$ for some $1leq p < infty$. Then, there exist functions $u_m in C^{infty}(bar{U})$ such that
begin{align*}
u_m rightarrow u quad textrm{ in } W^{k,p}(U)
end{align*}
$textbf{Question}$ Although we change the boundary condition like
begin{align*}
partial U=bigcup_{j=1}^n Gamma_j, quad (textrm{boundary is piecewise } C^{1})
end{align*}
where each $Gamma_j$ for $j=1, cdots, n$ is a $C^1$, $Gamma_j$ and $Gamma_{j^{'}}$ do not intersect except at their endpoints if $jneq j'$, then does the theorem still hold?
Any help is appreciated!!
I want to know references related that...
Thank you!!
analysis pde sobolev-spaces approximation-theory
asked Nov 14 at 9:19
w.sdka
32119
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In Evans,
$textbf{Theorem} $ (Global Approximation Theorem) Assume $U$ is bounded, and $partial U$ is $C^1$. Suppose as well that $u in W^{k,p}(U)$ for some $1leq p < infty$. Then, there exist functions $u_m in C^{infty}(bar{U})$ such that
begin{align*}
u_m rightarrow u quad textrm{ in } W^{k,p}(U)
end{align*}
$textbf{Question}$ Although we change the boundary condition like
begin{align*}
partial U=bigcup_{j=1}^n Gamma_j, quad (textrm{boundary is piecewise } C^{1})
end{align*}
where each $Gamma_j$ for $j=1, cdots, n$ is a $C^1$, $Gamma_j$ and $Gamma_{j^{'}}$ do not intersect except at their endpoints if $jneq j'$, then does the theorem still hold?
Any help is appreciated!!
I want to know references related that...
Thank you!!
analysis pde sobolev-spaces approximation-theory
asked Nov 14 at 9:19
w.sdka
32119
In Evans,
$textbf{Theorem} $ (Global Approximation Theorem) Assume $U$ is bounded, and $partial U$ is $C^1$. Suppose as well that $u in W^{k,p}(U)$ for some $1leq p < infty$. Then, there exist functions $u_m in C^{infty}(bar{U})$ such that
begin{align*}
u_m rightarrow u quad textrm{ in } W^{k,p}(U)
end{align*}
$textbf{Question}$ Although we change the boundary condition like
begin{align*}
partial U=bigcup_{j=1}^n Gamma_j, quad (textrm{boundary is piecewise } C^{1})
end{align*}
where each $Gamma_j$ for $j=1, cdots, n$ is a $C^1$, $Gamma_j$ and $Gamma_{j^{'}}$ do not intersect except at their endpoints if $jneq j'$, then does the theorem still hold?
Any help is appreciated!!
I want to know references related that...
Thank you!!
analysis pde sobolev-spaces approximation-theory
analysis pde sobolev-spaces approximation-theory
asked Nov 14 at 9:19
w.sdka
32119
asked Nov 14 at 9:19
w.sdka
32119
edited Nov 15 at 16:03
asked Nov 14 at 9:19
w.sdka
32119
asked Nov 14 at 9:19
w.sdka
32119
asked Nov 14 at 9:19
w.sdka
32119
32119
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
... does the theorem still hold?
Smooth boundary is not needed. The theorem is true provided that the domain does not lie on both sides of any part of its boundary, i.e., provided that $U$ satisfies the segment condition.
Reference: Adams book, p. 68.
answered Nov 26 at 15:31
Pedro
10.2k23066
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
... does the theorem still hold?
Smooth boundary is not needed. The theorem is true provided that the domain does not lie on both sides of any part of its boundary, i.e., provided that $U$ satisfies the segment condition.
Reference: Adams book, p. 68.
answered Nov 26 at 15:31
Pedro
10.2k23066
add a comment |
up vote
0
down vote
... does the theorem still hold?
Smooth boundary is not needed. The theorem is true provided that the domain does not lie on both sides of any part of its boundary, i.e., provided that $U$ satisfies the segment condition.
Reference: Adams book, p. 68.
answered Nov 26 at 15:31
Pedro
10.2k23066
add a comment |
up vote
0
down vote
up vote
0
down vote
... does the theorem still hold?
Smooth boundary is not needed. The theorem is true provided that the domain does not lie on both sides of any part of its boundary, i.e., provided that $U$ satisfies the segment condition.
Reference: Adams book, p. 68.
answered Nov 26 at 15:31
Pedro
10.2k23066
... does the theorem still hold?
Smooth boundary is not needed. The theorem is true provided that the domain does not lie on both sides of any part of its boundary, i.e., provided that $U$ satisfies the segment condition.
Reference: Adams book, p. 68.
answered Nov 26 at 15:31
Pedro
10.2k23066
edited Nov 26 at 16:11
answered Nov 26 at 15:31
Pedro
10.2k23066
answered Nov 26 at 15:31
Pedro
10.2k23066
answered Nov 26 at 15:31
Pedro
10.2k23066
10.2k23066
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2998025%2fextended-global-approximation-theorem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
