Do non-continuous Sobolev maps pull back closed forms to weakly closed forms?
$newcommand{R}{mathbb R}$
$newcommand{N}{mathbb N}$
$newcommand{de}{delta}$
$newcommand{sig}{sigma}$
$newcommand{Average}[1]{leftlangle#1rightrangle} $
$newcommand{IP}[2]{Average{#1,#2}}$
Let $n,d in mathbb{N}$, and let $Omega subseteq R^n$ be open.
Let $f in W^{1,k}(Omega,R^d)$. Let $omega in Omega^k(R^d)$ be a smooth closed $k$-form; Is it true that $f^*omega$ is weakly closed? i.e. does
$$int_{Omega} IP{f^* omega}{de sig}=0$$
hold for every compactly-supported $k+1$-form $sigma in Omega^{k+1}(Omega)$?
I have read that this should be true, but I am only able to show this in two special cases:
- The form $omega$ is constant.
$f$ is continuous.
Does this hold for non-continuous Sobolev maps in general?
Here is the problem as I see it: We approximate $f$ via smooth functions; suppose that $f_n in C^{infty}(Omega,R^d)$ satisfy $f_n to f$ in $W^{1,k}(Omega,R^d)$. Then
$$
int_{Omega} IP{f^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{ f_n^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{d f_n^* omega}{ sig}=0.
$$
Now, we need to justify the passage to the limit $int_{Omega} IP{f^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{ f_n^* omega}{de sig}$.
If $omega$ is constant, that is $omega_q= alpha$ independently of $q in R^d$, where $alpha$ is a fixed element in $bigwedge^k (R^d)^* $, then we have
$$
|f^* omega-f_n^* omega| le |alpha| , left|bigwedge^{k} df-bigwedge^{k} df_nright|_{op},
$$
thus
$$
left|int_{Omega} IP{f^* omega}{de sig}- IP{ f_n^* omega}{de sig}right| le |alpha| |de sig|_{sup} int_{Omega} left|bigwedge^{k} df-bigwedge^{k} df_nright|.
$$
and The RHS tends to zero since Sobolev approximation lifts to exterior powers.
When $omega$ is not constant, we have a problem that the point of evaluation "moves with the function" that pulls back, that is
$$
begin{split}
& |f^* omega-f_n^* omega|(p)= \
&left|omega_{f(p)} circ bigwedge^{k} df_p -omega_{f_n(p)} circ bigwedge^{k} (df_n)_p right| le \
&left|big(omega_{f(p)}-omega_{f_n(p)}) circ bigwedge^{k} df_p +omega_{f_n(p)} circ big( bigwedge^{k} df_p-bigwedge^{k} (df_n)_p) right| le \
&|omega_{f(p)}-omega_{f_n(p)}| , , cdot , , left| bigwedge^{k} df_pright| +|omega_{f_n(p)} | , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right|
end{split}
$$
So, we we need to estimate $omega_{f(p)}-omega_{f_n(p)}$. When $f$ is continuous, we can take $f_n$ which converges uniformly to $f$, and thus we can continue the estimate:
$$
begin{split}
&|omega_{f(p)}-omega_{f_n(p)}| , , cdot , , left| bigwedge^{k} df_pright| +|omega_{f_n(p)} | , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| le \
&|Tomega|_{sup} , cdot , |f(p)-f_n(p)| , cdot , left| bigwedge^{k} df_pright| +|omega |_{sup} , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| le \
&|Tomega|_{sup} , cdot , |f-f_n|_{sup} , cdot , left| bigwedge^{k} df_pright| +|omega |_{sup} , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| stackrel{L^1}{to} 0,
%&|alpha circ brk{bigwedge^{k} df_p-bigwedge^{k} (df_n)_p} | le |alpha| cdot |bigwedge^{k} df_p-bigwedge^{k} (df_n)_p|_{op}.
end{split}
$$
So, does the preservation of closedness hold in general (for non-continuous maps) or is there a counter-example?
dg.differential-geometry ap.analysis-of-pdes sobolev-spaces regularity exterior-algebra
add a comment |
$newcommand{R}{mathbb R}$
$newcommand{N}{mathbb N}$
$newcommand{de}{delta}$
$newcommand{sig}{sigma}$
$newcommand{Average}[1]{leftlangle#1rightrangle} $
$newcommand{IP}[2]{Average{#1,#2}}$
Let $n,d in mathbb{N}$, and let $Omega subseteq R^n$ be open.
Let $f in W^{1,k}(Omega,R^d)$. Let $omega in Omega^k(R^d)$ be a smooth closed $k$-form; Is it true that $f^*omega$ is weakly closed? i.e. does
$$int_{Omega} IP{f^* omega}{de sig}=0$$
hold for every compactly-supported $k+1$-form $sigma in Omega^{k+1}(Omega)$?
I have read that this should be true, but I am only able to show this in two special cases:
- The form $omega$ is constant.
$f$ is continuous.
Does this hold for non-continuous Sobolev maps in general?
Here is the problem as I see it: We approximate $f$ via smooth functions; suppose that $f_n in C^{infty}(Omega,R^d)$ satisfy $f_n to f$ in $W^{1,k}(Omega,R^d)$. Then
$$
int_{Omega} IP{f^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{ f_n^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{d f_n^* omega}{ sig}=0.
$$
Now, we need to justify the passage to the limit $int_{Omega} IP{f^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{ f_n^* omega}{de sig}$.
If $omega$ is constant, that is $omega_q= alpha$ independently of $q in R^d$, where $alpha$ is a fixed element in $bigwedge^k (R^d)^* $, then we have
$$
|f^* omega-f_n^* omega| le |alpha| , left|bigwedge^{k} df-bigwedge^{k} df_nright|_{op},
$$
thus
$$
left|int_{Omega} IP{f^* omega}{de sig}- IP{ f_n^* omega}{de sig}right| le |alpha| |de sig|_{sup} int_{Omega} left|bigwedge^{k} df-bigwedge^{k} df_nright|.
$$
and The RHS tends to zero since Sobolev approximation lifts to exterior powers.
When $omega$ is not constant, we have a problem that the point of evaluation "moves with the function" that pulls back, that is
$$
begin{split}
& |f^* omega-f_n^* omega|(p)= \
&left|omega_{f(p)} circ bigwedge^{k} df_p -omega_{f_n(p)} circ bigwedge^{k} (df_n)_p right| le \
&left|big(omega_{f(p)}-omega_{f_n(p)}) circ bigwedge^{k} df_p +omega_{f_n(p)} circ big( bigwedge^{k} df_p-bigwedge^{k} (df_n)_p) right| le \
&|omega_{f(p)}-omega_{f_n(p)}| , , cdot , , left| bigwedge^{k} df_pright| +|omega_{f_n(p)} | , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right|
end{split}
$$
So, we we need to estimate $omega_{f(p)}-omega_{f_n(p)}$. When $f$ is continuous, we can take $f_n$ which converges uniformly to $f$, and thus we can continue the estimate:
$$
begin{split}
&|omega_{f(p)}-omega_{f_n(p)}| , , cdot , , left| bigwedge^{k} df_pright| +|omega_{f_n(p)} | , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| le \
&|Tomega|_{sup} , cdot , |f(p)-f_n(p)| , cdot , left| bigwedge^{k} df_pright| +|omega |_{sup} , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| le \
&|Tomega|_{sup} , cdot , |f-f_n|_{sup} , cdot , left| bigwedge^{k} df_pright| +|omega |_{sup} , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| stackrel{L^1}{to} 0,
%&|alpha circ brk{bigwedge^{k} df_p-bigwedge^{k} (df_n)_p} | le |alpha| cdot |bigwedge^{k} df_p-bigwedge^{k} (df_n)_p|_{op}.
end{split}
$$
So, does the preservation of closedness hold in general (for non-continuous maps) or is there a counter-example?
dg.differential-geometry ap.analysis-of-pdes sobolev-spaces regularity exterior-algebra
add a comment |
$newcommand{R}{mathbb R}$
$newcommand{N}{mathbb N}$
$newcommand{de}{delta}$
$newcommand{sig}{sigma}$
$newcommand{Average}[1]{leftlangle#1rightrangle} $
$newcommand{IP}[2]{Average{#1,#2}}$
Let $n,d in mathbb{N}$, and let $Omega subseteq R^n$ be open.
Let $f in W^{1,k}(Omega,R^d)$. Let $omega in Omega^k(R^d)$ be a smooth closed $k$-form; Is it true that $f^*omega$ is weakly closed? i.e. does
$$int_{Omega} IP{f^* omega}{de sig}=0$$
hold for every compactly-supported $k+1$-form $sigma in Omega^{k+1}(Omega)$?
I have read that this should be true, but I am only able to show this in two special cases:
- The form $omega$ is constant.
$f$ is continuous.
Does this hold for non-continuous Sobolev maps in general?
Here is the problem as I see it: We approximate $f$ via smooth functions; suppose that $f_n in C^{infty}(Omega,R^d)$ satisfy $f_n to f$ in $W^{1,k}(Omega,R^d)$. Then
$$
int_{Omega} IP{f^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{ f_n^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{d f_n^* omega}{ sig}=0.
$$
Now, we need to justify the passage to the limit $int_{Omega} IP{f^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{ f_n^* omega}{de sig}$.
If $omega$ is constant, that is $omega_q= alpha$ independently of $q in R^d$, where $alpha$ is a fixed element in $bigwedge^k (R^d)^* $, then we have
$$
|f^* omega-f_n^* omega| le |alpha| , left|bigwedge^{k} df-bigwedge^{k} df_nright|_{op},
$$
thus
$$
left|int_{Omega} IP{f^* omega}{de sig}- IP{ f_n^* omega}{de sig}right| le |alpha| |de sig|_{sup} int_{Omega} left|bigwedge^{k} df-bigwedge^{k} df_nright|.
$$
and The RHS tends to zero since Sobolev approximation lifts to exterior powers.
When $omega$ is not constant, we have a problem that the point of evaluation "moves with the function" that pulls back, that is
$$
begin{split}
& |f^* omega-f_n^* omega|(p)= \
&left|omega_{f(p)} circ bigwedge^{k} df_p -omega_{f_n(p)} circ bigwedge^{k} (df_n)_p right| le \
&left|big(omega_{f(p)}-omega_{f_n(p)}) circ bigwedge^{k} df_p +omega_{f_n(p)} circ big( bigwedge^{k} df_p-bigwedge^{k} (df_n)_p) right| le \
&|omega_{f(p)}-omega_{f_n(p)}| , , cdot , , left| bigwedge^{k} df_pright| +|omega_{f_n(p)} | , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right|
end{split}
$$
So, we we need to estimate $omega_{f(p)}-omega_{f_n(p)}$. When $f$ is continuous, we can take $f_n$ which converges uniformly to $f$, and thus we can continue the estimate:
$$
begin{split}
&|omega_{f(p)}-omega_{f_n(p)}| , , cdot , , left| bigwedge^{k} df_pright| +|omega_{f_n(p)} | , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| le \
&|Tomega|_{sup} , cdot , |f(p)-f_n(p)| , cdot , left| bigwedge^{k} df_pright| +|omega |_{sup} , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| le \
&|Tomega|_{sup} , cdot , |f-f_n|_{sup} , cdot , left| bigwedge^{k} df_pright| +|omega |_{sup} , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| stackrel{L^1}{to} 0,
%&|alpha circ brk{bigwedge^{k} df_p-bigwedge^{k} (df_n)_p} | le |alpha| cdot |bigwedge^{k} df_p-bigwedge^{k} (df_n)_p|_{op}.
end{split}
$$
So, does the preservation of closedness hold in general (for non-continuous maps) or is there a counter-example?
dg.differential-geometry ap.analysis-of-pdes sobolev-spaces regularity exterior-algebra
$newcommand{R}{mathbb R}$
$newcommand{N}{mathbb N}$
$newcommand{de}{delta}$
$newcommand{sig}{sigma}$
$newcommand{Average}[1]{leftlangle#1rightrangle} $
$newcommand{IP}[2]{Average{#1,#2}}$
Let $n,d in mathbb{N}$, and let $Omega subseteq R^n$ be open.
Let $f in W^{1,k}(Omega,R^d)$. Let $omega in Omega^k(R^d)$ be a smooth closed $k$-form; Is it true that $f^*omega$ is weakly closed? i.e. does
$$int_{Omega} IP{f^* omega}{de sig}=0$$
hold for every compactly-supported $k+1$-form $sigma in Omega^{k+1}(Omega)$?
I have read that this should be true, but I am only able to show this in two special cases:
- The form $omega$ is constant.
$f$ is continuous.
Does this hold for non-continuous Sobolev maps in general?
Here is the problem as I see it: We approximate $f$ via smooth functions; suppose that $f_n in C^{infty}(Omega,R^d)$ satisfy $f_n to f$ in $W^{1,k}(Omega,R^d)$. Then
$$
int_{Omega} IP{f^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{ f_n^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{d f_n^* omega}{ sig}=0.
$$
Now, we need to justify the passage to the limit $int_{Omega} IP{f^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{ f_n^* omega}{de sig}$.
If $omega$ is constant, that is $omega_q= alpha$ independently of $q in R^d$, where $alpha$ is a fixed element in $bigwedge^k (R^d)^* $, then we have
$$
|f^* omega-f_n^* omega| le |alpha| , left|bigwedge^{k} df-bigwedge^{k} df_nright|_{op},
$$
thus
$$
left|int_{Omega} IP{f^* omega}{de sig}- IP{ f_n^* omega}{de sig}right| le |alpha| |de sig|_{sup} int_{Omega} left|bigwedge^{k} df-bigwedge^{k} df_nright|.
$$
and The RHS tends to zero since Sobolev approximation lifts to exterior powers.
When $omega$ is not constant, we have a problem that the point of evaluation "moves with the function" that pulls back, that is
$$
begin{split}
& |f^* omega-f_n^* omega|(p)= \
&left|omega_{f(p)} circ bigwedge^{k} df_p -omega_{f_n(p)} circ bigwedge^{k} (df_n)_p right| le \
&left|big(omega_{f(p)}-omega_{f_n(p)}) circ bigwedge^{k} df_p +omega_{f_n(p)} circ big( bigwedge^{k} df_p-bigwedge^{k} (df_n)_p) right| le \
&|omega_{f(p)}-omega_{f_n(p)}| , , cdot , , left| bigwedge^{k} df_pright| +|omega_{f_n(p)} | , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right|
end{split}
$$
So, we we need to estimate $omega_{f(p)}-omega_{f_n(p)}$. When $f$ is continuous, we can take $f_n$ which converges uniformly to $f$, and thus we can continue the estimate:
$$
begin{split}
&|omega_{f(p)}-omega_{f_n(p)}| , , cdot , , left| bigwedge^{k} df_pright| +|omega_{f_n(p)} | , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| le \
&|Tomega|_{sup} , cdot , |f(p)-f_n(p)| , cdot , left| bigwedge^{k} df_pright| +|omega |_{sup} , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| le \
&|Tomega|_{sup} , cdot , |f-f_n|_{sup} , cdot , left| bigwedge^{k} df_pright| +|omega |_{sup} , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| stackrel{L^1}{to} 0,
%&|alpha circ brk{bigwedge^{k} df_p-bigwedge^{k} (df_n)_p} | le |alpha| cdot |bigwedge^{k} df_p-bigwedge^{k} (df_n)_p|_{op}.
end{split}
$$
So, does the preservation of closedness hold in general (for non-continuous maps) or is there a counter-example?
dg.differential-geometry ap.analysis-of-pdes sobolev-spaces regularity exterior-algebra
dg.differential-geometry ap.analysis-of-pdes sobolev-spaces regularity exterior-algebra
asked 2 hours ago
Asaf ShacharAsaf Shachar
2,2391746
2,2391746
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
This is I guess Lemma 4.1 in https://arxiv.org/pdf/1301.4978.pdf. The assumptions might be a bit different than yours, but it might be useful. I guess the assumptions in Lemma 4.1 are close to be optimal. I will expand my answer when I have time. Let me know if Lemma 4.1 suits you.
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "504"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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%2fmathoverflow.net%2fquestions%2f320789%2fdo-non-continuous-sobolev-maps-pull-back-closed-forms-to-weakly-closed-forms%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is I guess Lemma 4.1 in https://arxiv.org/pdf/1301.4978.pdf. The assumptions might be a bit different than yours, but it might be useful. I guess the assumptions in Lemma 4.1 are close to be optimal. I will expand my answer when I have time. Let me know if Lemma 4.1 suits you.
add a comment |
This is I guess Lemma 4.1 in https://arxiv.org/pdf/1301.4978.pdf. The assumptions might be a bit different than yours, but it might be useful. I guess the assumptions in Lemma 4.1 are close to be optimal. I will expand my answer when I have time. Let me know if Lemma 4.1 suits you.
add a comment |
This is I guess Lemma 4.1 in https://arxiv.org/pdf/1301.4978.pdf. The assumptions might be a bit different than yours, but it might be useful. I guess the assumptions in Lemma 4.1 are close to be optimal. I will expand my answer when I have time. Let me know if Lemma 4.1 suits you.
This is I guess Lemma 4.1 in https://arxiv.org/pdf/1301.4978.pdf. The assumptions might be a bit different than yours, but it might be useful. I guess the assumptions in Lemma 4.1 are close to be optimal. I will expand my answer when I have time. Let me know if Lemma 4.1 suits you.
edited 42 mins ago
Martin Sleziak
2,95532028
2,95532028
answered 47 mins ago
Piotr HajlaszPiotr Hajlasz
6,50442356
6,50442356
add a comment |
add a comment |
Thanks for contributing an answer to MathOverflow!
- 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.
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%2fmathoverflow.net%2fquestions%2f320789%2fdo-non-continuous-sobolev-maps-pull-back-closed-forms-to-weakly-closed-forms%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