Mixing
A direct proof of the surjectivity of divergence operator
$begingroup$
The main question that I want to solve is that:
Let ${X^k}_{k=-infty}^infty$ be a sequence of (Banach) spaces which are subsets of distribution space $mathcal D'(Omega)$ where $Omegasubsetmathbb R^n$, satisfies
(a) $fin X^k$ iff $nabla fin X^{k-1}otimesmathbb R^n$;
(b) Either (i) $nabla$ is injective, or (ii) polynomials lays in $bigcap_{k=1}^infty X^k$;
Then the divergence $operatorname{div}:X^kotimesmathbb R^nto X^{k-1}$ is surjection.
The answer in Show that the divergence operator is onto gives a proof of the surjection of $operatorname{div}:L^2(mathbb R^n)to H^1(mathbb R^n)$, moreover the same argument shows is also holds if $X^k$ are reflexive spaces. But it use Hahn-Banach which is non-constructive.
Here is the idea that I already have, denote $Y^k={finmathcal D'(Omega):(phimapstolangle f,phirangle)in(X^k)^*}$, then ${Y^k}$ satisfies (a) and (b) ($Y^k$ satisfies (b) maybe a bit nontrivial).
Set $tilde X^k={finmathcal D'(Omega):(phimapstolangle f,phirangle)in(Y^k)^*}$. Then $X^ksubsettilde X^k$ is a closed subspace, and the argument shows that given $fin X^k$ there a preimage $Fintilde X^{k+1}otimesmathbb R^n$ satisfies $f=operatorname{div} F$.
If $tilde X^k=X^k$ the proof is done (inparticular $X^k=C_b^k(Omega)$ or $X^k=W^{k,p}(Omega)$ are true). But what if $tilde X$ is strictly larger? The non-constructive can only guarantee the existence in $tilde X^{k+1}$ instead of $ X^{k+1}$, and I didn't find a good characterization for the kernel of divergence.
functional-analysis distribution-theory divergence
asked Jan 6 at 20:48
yaolidingyaoliding
567312
$endgroup$
add a comment |
$begingroup$
The main question that I want to solve is that:
Let ${X^k}_{k=-infty}^infty$ be a sequence of (Banach) spaces which are subsets of distribution space $mathcal D'(Omega)$ where $Omegasubsetmathbb R^n$, satisfies
(a) $fin X^k$ iff $nabla fin X^{k-1}otimesmathbb R^n$;
(b) Either (i) $nabla$ is injective, or (ii) polynomials lays in $bigcap_{k=1}^infty X^k$;
Then the divergence $operatorname{div}:X^kotimesmathbb R^nto X^{k-1}$ is surjection.
The answer in Show that the divergence operator is onto gives a proof of the surjection of $operatorname{div}:L^2(mathbb R^n)to H^1(mathbb R^n)$, moreover the same argument shows is also holds if $X^k$ are reflexive spaces. But it use Hahn-Banach which is non-constructive.
Here is the idea that I already have, denote $Y^k={finmathcal D'(Omega):(phimapstolangle f,phirangle)in(X^k)^*}$, then ${Y^k}$ satisfies (a) and (b) ($Y^k$ satisfies (b) maybe a bit nontrivial).
Set $tilde X^k={finmathcal D'(Omega):(phimapstolangle f,phirangle)in(Y^k)^*}$. Then $X^ksubsettilde X^k$ is a closed subspace, and the argument shows that given $fin X^k$ there a preimage $Fintilde X^{k+1}otimesmathbb R^n$ satisfies $f=operatorname{div} F$.
If $tilde X^k=X^k$ the proof is done (inparticular $X^k=C_b^k(Omega)$ or $X^k=W^{k,p}(Omega)$ are true). But what if $tilde X$ is strictly larger? The non-constructive can only guarantee the existence in $tilde X^{k+1}$ instead of $ X^{k+1}$, and I didn't find a good characterization for the kernel of divergence.
functional-analysis distribution-theory divergence
asked Jan 6 at 20:48
yaolidingyaoliding
567312
$endgroup$
add a comment |
$begingroup$
The main question that I want to solve is that:
Let ${X^k}_{k=-infty}^infty$ be a sequence of (Banach) spaces which are subsets of distribution space $mathcal D'(Omega)$ where $Omegasubsetmathbb R^n$, satisfies
(a) $fin X^k$ iff $nabla fin X^{k-1}otimesmathbb R^n$;
(b) Either (i) $nabla$ is injective, or (ii) polynomials lays in $bigcap_{k=1}^infty X^k$;
Then the divergence $operatorname{div}:X^kotimesmathbb R^nto X^{k-1}$ is surjection.
The answer in Show that the divergence operator is onto gives a proof of the surjection of $operatorname{div}:L^2(mathbb R^n)to H^1(mathbb R^n)$, moreover the same argument shows is also holds if $X^k$ are reflexive spaces. But it use Hahn-Banach which is non-constructive.
Here is the idea that I already have, denote $Y^k={finmathcal D'(Omega):(phimapstolangle f,phirangle)in(X^k)^*}$, then ${Y^k}$ satisfies (a) and (b) ($Y^k$ satisfies (b) maybe a bit nontrivial).
Set $tilde X^k={finmathcal D'(Omega):(phimapstolangle f,phirangle)in(Y^k)^*}$. Then $X^ksubsettilde X^k$ is a closed subspace, and the argument shows that given $fin X^k$ there a preimage $Fintilde X^{k+1}otimesmathbb R^n$ satisfies $f=operatorname{div} F$.
If $tilde X^k=X^k$ the proof is done (inparticular $X^k=C_b^k(Omega)$ or $X^k=W^{k,p}(Omega)$ are true). But what if $tilde X$ is strictly larger? The non-constructive can only guarantee the existence in $tilde X^{k+1}$ instead of $ X^{k+1}$, and I didn't find a good characterization for the kernel of divergence.
functional-analysis distribution-theory divergence
asked Jan 6 at 20:48
yaolidingyaoliding
567312
$endgroup$
The main question that I want to solve is that:
Let ${X^k}_{k=-infty}^infty$ be a sequence of (Banach) spaces which are subsets of distribution space $mathcal D'(Omega)$ where $Omegasubsetmathbb R^n$, satisfies
(a) $fin X^k$ iff $nabla fin X^{k-1}otimesmathbb R^n$;
(b) Either (i) $nabla$ is injective, or (ii) polynomials lays in $bigcap_{k=1}^infty X^k$;
Then the divergence $operatorname{div}:X^kotimesmathbb R^nto X^{k-1}$ is surjection.
The answer in Show that the divergence operator is onto gives a proof of the surjection of $operatorname{div}:L^2(mathbb R^n)to H^1(mathbb R^n)$, moreover the same argument shows is also holds if $X^k$ are reflexive spaces. But it use Hahn-Banach which is non-constructive.
Here is the idea that I already have, denote $Y^k={finmathcal D'(Omega):(phimapstolangle f,phirangle)in(X^k)^*}$, then ${Y^k}$ satisfies (a) and (b) ($Y^k$ satisfies (b) maybe a bit nontrivial).
Set $tilde X^k={finmathcal D'(Omega):(phimapstolangle f,phirangle)in(Y^k)^*}$. Then $X^ksubsettilde X^k$ is a closed subspace, and the argument shows that given $fin X^k$ there a preimage $Fintilde X^{k+1}otimesmathbb R^n$ satisfies $f=operatorname{div} F$.
If $tilde X^k=X^k$ the proof is done (inparticular $X^k=C_b^k(Omega)$ or $X^k=W^{k,p}(Omega)$ are true). But what if $tilde X$ is strictly larger? The non-constructive can only guarantee the existence in $tilde X^{k+1}$ instead of $ X^{k+1}$, and I didn't find a good characterization for the kernel of divergence.
functional-analysis distribution-theory divergence
functional-analysis distribution-theory divergence
asked Jan 6 at 20:48
yaolidingyaoliding
567312
asked Jan 6 at 20:48
yaolidingyaoliding
567312
asked Jan 6 at 20:48
yaolidingyaoliding
567312
asked Jan 6 at 20:48
yaolidingyaoliding
567312
asked Jan 6 at 20:48
yaolidingyaoliding
567312
567312
add a comment |
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
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%2fmath.stackexchange.com%2fquestions%2f3064376%2fa-direct-proof-of-the-surjectivity-of-divergence-operator%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
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%2f3064376%2fa-direct-proof-of-the-surjectivity-of-divergence-operator%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
