Question on Sobolev extension onto boundary
Let $U subset mathbb R^3$ be an open, bounded and connected set with a $C^2-$regular boundary $partial U$. I'm trying to understand the following implication:
If $fin W^{2-1/2,2}(U)$ then $f{vert}_{partial U} in W^{1,2}(partial U)(*)$
So, I'm aware of this theorem:
General Trace Theorem: if $fin W^{1-1/p,p}(partial Omega)$, then there exists a function $f in W^{1,p}(Omega)$ such that $f{vert}_{partial Omega}=f$
QUESTION: Is the above theorem still valid if we replace $partial Omega$ with $U$ and $Omega$ with $partial U$ so that $(*)$ makes sense? If not, is there any other way to deduce $(*)$?
Any help is appreciated. Thanks in advance!
EDIT: regularity in $(*)$ fixed
functional-analysis pde sobolev-spaces trace regularity-theory-of-pdes
asked Nov 13 at 15:51
kaithkolesidou
959511
|
show 3 more comments
Let $U subset mathbb R^3$ be an open, bounded and connected set with a $C^2-$regular boundary $partial U$. I'm trying to understand the following implication:
If $fin W^{2-1/2,2}(U)$ then $f{vert}_{partial U} in W^{1,2}(partial U)(*)$
So, I'm aware of this theorem:
General Trace Theorem: if $fin W^{1-1/p,p}(partial Omega)$, then there exists a function $f in W^{1,p}(Omega)$ such that $f{vert}_{partial Omega}=f$
QUESTION: Is the above theorem still valid if we replace $partial Omega$ with $U$ and $Omega$ with $partial U$ so that $(*)$ makes sense? If not, is there any other way to deduce $(*)$?
Any help is appreciated. Thanks in advance!
EDIT: regularity in $(*)$ fixed
functional-analysis pde sobolev-spaces trace regularity-theory-of-pdes
asked Nov 13 at 15:51
kaithkolesidou
959511
You're asking about whether you can extend a function from the boundary into the domain?
– Ian
Nov 13 at 15:52
I think that you have the regularities reversed. Anyway, see here.
– Giuseppe Negro
Nov 13 at 15:55
@Ian no... I am trying to understand the implication in $(*)$. There the extension is from the domain to boundary...
– kaithkolesidou
Nov 13 at 15:56
The given theorem is about extending from the inside of the domain to the boundary. Switching the two around would amount to extending a function given on the boundary to the domain, unless I am misunderstanding your intended meaning.
– Ian
Nov 13 at 15:57
@GiuseppeNegro What do you mean by reversed regularities?
– kaithkolesidou
Nov 13 at 16:00
|
show 3 more comments
Let $U subset mathbb R^3$ be an open, bounded and connected set with a $C^2-$regular boundary $partial U$. I'm trying to understand the following implication:
If $fin W^{2-1/2,2}(U)$ then $f{vert}_{partial U} in W^{1,2}(partial U)(*)$
So, I'm aware of this theorem:
General Trace Theorem: if $fin W^{1-1/p,p}(partial Omega)$, then there exists a function $f in W^{1,p}(Omega)$ such that $f{vert}_{partial Omega}=f$
QUESTION: Is the above theorem still valid if we replace $partial Omega$ with $U$ and $Omega$ with $partial U$ so that $(*)$ makes sense? If not, is there any other way to deduce $(*)$?
Any help is appreciated. Thanks in advance!
EDIT: regularity in $(*)$ fixed
functional-analysis pde sobolev-spaces trace regularity-theory-of-pdes
asked Nov 13 at 15:51
kaithkolesidou
959511
Let $U subset mathbb R^3$ be an open, bounded and connected set with a $C^2-$regular boundary $partial U$. I'm trying to understand the following implication:
If $fin W^{2-1/2,2}(U)$ then $f{vert}_{partial U} in W^{1,2}(partial U)(*)$
So, I'm aware of this theorem:
General Trace Theorem: if $fin W^{1-1/p,p}(partial Omega)$, then there exists a function $f in W^{1,p}(Omega)$ such that $f{vert}_{partial Omega}=f$
QUESTION: Is the above theorem still valid if we replace $partial Omega$ with $U$ and $Omega$ with $partial U$ so that $(*)$ makes sense? If not, is there any other way to deduce $(*)$?
Any help is appreciated. Thanks in advance!
EDIT: regularity in $(*)$ fixed
functional-analysis pde sobolev-spaces trace regularity-theory-of-pdes
functional-analysis pde sobolev-spaces trace regularity-theory-of-pdes
asked Nov 13 at 15:51
kaithkolesidou
959511
asked Nov 13 at 15:51
kaithkolesidou
959511
edited Nov 29 at 13:25
asked Nov 13 at 15:51
kaithkolesidou
959511
asked Nov 13 at 15:51
kaithkolesidou
959511
asked Nov 13 at 15:51
kaithkolesidou
959511
959511
You're asking about whether you can extend a function from the boundary into the domain?
– Ian
Nov 13 at 15:52
I think that you have the regularities reversed. Anyway, see here.
– Giuseppe Negro
Nov 13 at 15:55
@Ian no... I am trying to understand the implication in $(*)$. There the extension is from the domain to boundary...
– kaithkolesidou
Nov 13 at 15:56
The given theorem is about extending from the inside of the domain to the boundary. Switching the two around would amount to extending a function given on the boundary to the domain, unless I am misunderstanding your intended meaning.
– Ian
Nov 13 at 15:57
@GiuseppeNegro What do you mean by reversed regularities?
– kaithkolesidou
Nov 13 at 16:00
|
show 3 more comments
You're asking about whether you can extend a function from the boundary into the domain?
– Ian
Nov 13 at 15:52
I think that you have the regularities reversed. Anyway, see here.
– Giuseppe Negro
Nov 13 at 15:55
@Ian no... I am trying to understand the implication in $(*)$. There the extension is from the domain to boundary...
– kaithkolesidou
Nov 13 at 15:56
The given theorem is about extending from the inside of the domain to the boundary. Switching the two around would amount to extending a function given on the boundary to the domain, unless I am misunderstanding your intended meaning.
– Ian
Nov 13 at 15:57
@GiuseppeNegro What do you mean by reversed regularities?
– kaithkolesidou
Nov 13 at 16:00
You're asking about whether you can extend a function from the boundary into the domain?
– Ian
Nov 13 at 15:52
You're asking about whether you can extend a function from the boundary into the domain?
– Ian
Nov 13 at 15:52
I think that you have the regularities reversed. Anyway, see here.
– Giuseppe Negro
Nov 13 at 15:55
I think that you have the regularities reversed. Anyway, see here.
– Giuseppe Negro
Nov 13 at 15:55
@Ian no... I am trying to understand the implication in $(*)$. There the extension is from the domain to boundary...
– kaithkolesidou
Nov 13 at 15:56
@Ian no... I am trying to understand the implication in $(*)$. There the extension is from the domain to boundary...
– kaithkolesidou
Nov 13 at 15:56
The given theorem is about extending from the inside of the domain to the boundary. Switching the two around would amount to extending a function given on the boundary to the domain, unless I am misunderstanding your intended meaning.
– Ian
Nov 13 at 15:57
The given theorem is about extending from the inside of the domain to the boundary. Switching the two around would amount to extending a function given on the boundary to the domain, unless I am misunderstanding your intended meaning.
– Ian
Nov 13 at 15:57
@GiuseppeNegro What do you mean by reversed regularities?
– kaithkolesidou
Nov 13 at 16:00
@GiuseppeNegro What do you mean by reversed regularities?
– kaithkolesidou
Nov 13 at 16:00
|
show 3 more comments
1 Answer
1
active
oldest
votes
$newcommand{R}{mathbb{R}}$
The implication (*) is invalid, and for obvious reasons. The trace $f|_{partial U}$ is kind of a restriction of $f$, so you cannot expect it to have more derivatives than $f$!
To have an example, choose your favorite function $g in W^{1/2,2}(R)$ that doesn't belong to $W^{1,2}(R)$. Then consider
begin{align*}
U &= { (x,y,z) in R^3 : z > 0 }, \
f(x,y,z) & = g(x,y).
end{align*}
Then $f$ is locally in $W^{1/2,2}$, but its restriction to the boundary $partial U = R^2 times {0}$ is simply $g$, which is not in $W^{1,2}$.
Of course, one can modify this example to have a bounded domain.
answered Nov 13 at 18:51
Michał Miśkiewicz
2,743616
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
votes
$newcommand{R}{mathbb{R}}$
The implication (*) is invalid, and for obvious reasons. The trace $f|_{partial U}$ is kind of a restriction of $f$, so you cannot expect it to have more derivatives than $f$!
To have an example, choose your favorite function $g in W^{1/2,2}(R)$ that doesn't belong to $W^{1,2}(R)$. Then consider
begin{align*}
U &= { (x,y,z) in R^3 : z > 0 }, \
f(x,y,z) & = g(x,y).
end{align*}
Then $f$ is locally in $W^{1/2,2}$, but its restriction to the boundary $partial U = R^2 times {0}$ is simply $g$, which is not in $W^{1,2}$.
Of course, one can modify this example to have a bounded domain.
answered Nov 13 at 18:51
Michał Miśkiewicz
2,743616
add a comment |
$newcommand{R}{mathbb{R}}$
The implication (*) is invalid, and for obvious reasons. The trace $f|_{partial U}$ is kind of a restriction of $f$, so you cannot expect it to have more derivatives than $f$!
To have an example, choose your favorite function $g in W^{1/2,2}(R)$ that doesn't belong to $W^{1,2}(R)$. Then consider
begin{align*}
U &= { (x,y,z) in R^3 : z > 0 }, \
f(x,y,z) & = g(x,y).
end{align*}
Then $f$ is locally in $W^{1/2,2}$, but its restriction to the boundary $partial U = R^2 times {0}$ is simply $g$, which is not in $W^{1,2}$.
Of course, one can modify this example to have a bounded domain.
answered Nov 13 at 18:51
Michał Miśkiewicz
2,743616
add a comment |
$newcommand{R}{mathbb{R}}$
The implication (*) is invalid, and for obvious reasons. The trace $f|_{partial U}$ is kind of a restriction of $f$, so you cannot expect it to have more derivatives than $f$!
To have an example, choose your favorite function $g in W^{1/2,2}(R)$ that doesn't belong to $W^{1,2}(R)$. Then consider
begin{align*}
U &= { (x,y,z) in R^3 : z > 0 }, \
f(x,y,z) & = g(x,y).
end{align*}
Then $f$ is locally in $W^{1/2,2}$, but its restriction to the boundary $partial U = R^2 times {0}$ is simply $g$, which is not in $W^{1,2}$.
Of course, one can modify this example to have a bounded domain.
answered Nov 13 at 18:51
Michał Miśkiewicz
2,743616
$newcommand{R}{mathbb{R}}$
The implication (*) is invalid, and for obvious reasons. The trace $f|_{partial U}$ is kind of a restriction of $f$, so you cannot expect it to have more derivatives than $f$!
To have an example, choose your favorite function $g in W^{1/2,2}(R)$ that doesn't belong to $W^{1,2}(R)$. Then consider
begin{align*}
U &= { (x,y,z) in R^3 : z > 0 }, \
f(x,y,z) & = g(x,y).
end{align*}
Then $f$ is locally in $W^{1/2,2}$, but its restriction to the boundary $partial U = R^2 times {0}$ is simply $g$, which is not in $W^{1,2}$.
Of course, one can modify this example to have a bounded domain.
answered Nov 13 at 18:51
Michał Miśkiewicz
2,743616
answered Nov 13 at 18:51
Michał Miśkiewicz
2,743616
answered Nov 13 at 18:51
Michał Miśkiewicz
2,743616
answered Nov 13 at 18:51
Michał Miśkiewicz
2,743616
2,743616
add a comment |
add a comment |
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You're asking about whether you can extend a function from the boundary into the domain?
– Ian
Nov 13 at 15:52
I think that you have the regularities reversed. Anyway, see here.
– Giuseppe Negro
Nov 13 at 15:55
@Ian no... I am trying to understand the implication in $(*)$. There the extension is from the domain to boundary...
– kaithkolesidou
Nov 13 at 15:56
The given theorem is about extending from the inside of the domain to the boundary. Switching the two around would amount to extending a function given on the boundary to the domain, unless I am misunderstanding your intended meaning.
– Ian
Nov 13 at 15:57
@GiuseppeNegro What do you mean by reversed regularities?
– kaithkolesidou
Nov 13 at 16:00