Poincare Inequality implies Equivalent Norms
I am currently working through the subject of Sobolev Spaces using the book 'Partial Differential Equations' by Lawrence Evans. After the result proving the Poincare Inequality it says the following in the book(page 266.) "In view of the Poincare Inequality, on $W_{0}^{1,p}(U)$ the norm $||DU||_{L^{p}}$ is equivalent to $||u||_{W^{1,p}(U)}$, if $U$ is bounded." Do you know the argument behind this statement?
functional-analysis inequality pde sobolev-spaces norm
asked Oct 22 '13 at 10:12
Lucio DLucio D
351419
add a comment |
I am currently working through the subject of Sobolev Spaces using the book 'Partial Differential Equations' by Lawrence Evans. After the result proving the Poincare Inequality it says the following in the book(page 266.) "In view of the Poincare Inequality, on $W_{0}^{1,p}(U)$ the norm $||DU||_{L^{p}}$ is equivalent to $||u||_{W^{1,p}(U)}$, if $U$ is bounded." Do you know the argument behind this statement?
functional-analysis inequality pde sobolev-spaces norm
asked Oct 22 '13 at 10:12
Lucio DLucio D
351419
I can't understand why do you accepted Alex answers if he gives you only a partial answer (also to complicated).
– Tomás
Oct 23 '13 at 18:30
I don't understand why you think it is a partial answer? It seems fine to me. He is doing it using results from Evans book.
– Lucio D
Oct 23 '13 at 18:38
He considered only the case $1leq p< n$, while Julian proof is valid for all cases.
– Tomás
Oct 23 '13 at 19:02
Yes I agree Julien's is simpler and stronger answer seeing as it is more general. The Remark in Evans is referring to that particular case of $1 leq p < n$. I just ticked Alex's because he used the proofs presented in Evans up until that point, so it followed well in the sequence in which I am studying it. But I indicated 'useful answer for both'. I think Julien made a type though, it should be $||Du||_{L^{p}(U)} leq ||Du||_{L^{p}(U)} + ||u||_{L^{p}(U)} = ||u||_{W_{0}^{1,p}(U)} leq ||Du||_{L^{p}(U)} + C||Du||_{L^{p}(U)}$.
– Lucio D
Oct 23 '13 at 19:16
Ok, in the end you are aware that the inequality is valid for all $p$.
– Tomás
Oct 23 '13 at 19:19
add a comment |
I am currently working through the subject of Sobolev Spaces using the book 'Partial Differential Equations' by Lawrence Evans. After the result proving the Poincare Inequality it says the following in the book(page 266.) "In view of the Poincare Inequality, on $W_{0}^{1,p}(U)$ the norm $||DU||_{L^{p}}$ is equivalent to $||u||_{W^{1,p}(U)}$, if $U$ is bounded." Do you know the argument behind this statement?
functional-analysis inequality pde sobolev-spaces norm
asked Oct 22 '13 at 10:12
Lucio DLucio D
351419
I am currently working through the subject of Sobolev Spaces using the book 'Partial Differential Equations' by Lawrence Evans. After the result proving the Poincare Inequality it says the following in the book(page 266.) "In view of the Poincare Inequality, on $W_{0}^{1,p}(U)$ the norm $||DU||_{L^{p}}$ is equivalent to $||u||_{W^{1,p}(U)}$, if $U$ is bounded." Do you know the argument behind this statement?
functional-analysis inequality pde sobolev-spaces norm
functional-analysis inequality pde sobolev-spaces norm
asked Oct 22 '13 at 10:12
Lucio DLucio D
351419
asked Oct 22 '13 at 10:12
Lucio DLucio D
351419
asked Oct 22 '13 at 10:12
Lucio DLucio D
351419
asked Oct 22 '13 at 10:12
Lucio DLucio D
351419
asked Oct 22 '13 at 10:12
Lucio DLucio D
351419
351419
I can't understand why do you accepted Alex answers if he gives you only a partial answer (also to complicated).
– Tomás
Oct 23 '13 at 18:30
I don't understand why you think it is a partial answer? It seems fine to me. He is doing it using results from Evans book.
– Lucio D
Oct 23 '13 at 18:38
He considered only the case $1leq p< n$, while Julian proof is valid for all cases.
– Tomás
Oct 23 '13 at 19:02
Yes I agree Julien's is simpler and stronger answer seeing as it is more general. The Remark in Evans is referring to that particular case of $1 leq p < n$. I just ticked Alex's because he used the proofs presented in Evans up until that point, so it followed well in the sequence in which I am studying it. But I indicated 'useful answer for both'. I think Julien made a type though, it should be $||Du||_{L^{p}(U)} leq ||Du||_{L^{p}(U)} + ||u||_{L^{p}(U)} = ||u||_{W_{0}^{1,p}(U)} leq ||Du||_{L^{p}(U)} + C||Du||_{L^{p}(U)}$.
– Lucio D
Oct 23 '13 at 19:16
Ok, in the end you are aware that the inequality is valid for all $p$.
– Tomás
Oct 23 '13 at 19:19
add a comment |
I can't understand why do you accepted Alex answers if he gives you only a partial answer (also to complicated).
– Tomás
Oct 23 '13 at 18:30
I don't understand why you think it is a partial answer? It seems fine to me. He is doing it using results from Evans book.
– Lucio D
Oct 23 '13 at 18:38
He considered only the case $1leq p< n$, while Julian proof is valid for all cases.
– Tomás
Oct 23 '13 at 19:02
Yes I agree Julien's is simpler and stronger answer seeing as it is more general. The Remark in Evans is referring to that particular case of $1 leq p < n$. I just ticked Alex's because he used the proofs presented in Evans up until that point, so it followed well in the sequence in which I am studying it. But I indicated 'useful answer for both'. I think Julien made a type though, it should be $||Du||_{L^{p}(U)} leq ||Du||_{L^{p}(U)} + ||u||_{L^{p}(U)} = ||u||_{W_{0}^{1,p}(U)} leq ||Du||_{L^{p}(U)} + C||Du||_{L^{p}(U)}$.
– Lucio D
Oct 23 '13 at 19:16
Ok, in the end you are aware that the inequality is valid for all $p$.
– Tomás
Oct 23 '13 at 19:19
I can't understand why do you accepted Alex answers if he gives you only a partial answer (also to complicated).
– Tomás
Oct 23 '13 at 18:30
I can't understand why do you accepted Alex answers if he gives you only a partial answer (also to complicated).
– Tomás
Oct 23 '13 at 18:30
I don't understand why you think it is a partial answer? It seems fine to me. He is doing it using results from Evans book.
– Lucio D
Oct 23 '13 at 18:38
I don't understand why you think it is a partial answer? It seems fine to me. He is doing it using results from Evans book.
– Lucio D
Oct 23 '13 at 18:38
He considered only the case $1leq p< n$, while Julian proof is valid for all cases.
– Tomás
Oct 23 '13 at 19:02
He considered only the case $1leq p< n$, while Julian proof is valid for all cases.
– Tomás
Oct 23 '13 at 19:02
Yes I agree Julien's is simpler and stronger answer seeing as it is more general. The Remark in Evans is referring to that particular case of $1 leq p < n$. I just ticked Alex's because he used the proofs presented in Evans up until that point, so it followed well in the sequence in which I am studying it. But I indicated 'useful answer for both'. I think Julien made a type though, it should be $||Du||_{L^{p}(U)} leq ||Du||_{L^{p}(U)} + ||u||_{L^{p}(U)} = ||u||_{W_{0}^{1,p}(U)} leq ||Du||_{L^{p}(U)} + C||Du||_{L^{p}(U)}$.
– Lucio D
Oct 23 '13 at 19:16
Yes I agree Julien's is simpler and stronger answer seeing as it is more general. The Remark in Evans is referring to that particular case of $1 leq p < n$. I just ticked Alex's because he used the proofs presented in Evans up until that point, so it followed well in the sequence in which I am studying it. But I indicated 'useful answer for both'. I think Julien made a type though, it should be $||Du||_{L^{p}(U)} leq ||Du||_{L^{p}(U)} + ||u||_{L^{p}(U)} = ||u||_{W_{0}^{1,p}(U)} leq ||Du||_{L^{p}(U)} + C||Du||_{L^{p}(U)}$.
– Lucio D
Oct 23 '13 at 19:16
Ok, in the end you are aware that the inequality is valid for all $p$.
– Tomás
Oct 23 '13 at 19:19
Ok, in the end you are aware that the inequality is valid for all $p$.
– Tomás
Oct 23 '13 at 19:19
add a comment |
2 Answers
2
active
oldest
votes
This applies to $1 leq p < n$. You want constants $B_{1}$ and $B_{2}$ such that $B_{1}||u||_{W_{0}^{1,p}(U)} leq ||Du||_{L^{p}(U)} leq B_{2}||u||_{W_{0}^{1,p}}$. The inequality $||Du||_{L^{p}(U)} leq B_{2}||u||_{W_{0}^{1,p}}$ is trivial.
From Gagliardo-Nirenberg-Sobolev Inequality we have:
$||u||_{L^{p*}(U)} leq C||Du||_{L^{p}(U)}$ where $p* := frac{np}{n-p}$ $p* > p$.
We also have $||u||_{L^{q}(U)} leq C||u||_{L^{p*}(U)}$ if $1leq q leq p*$ by Generalized Holder Inequality.
If we take $A = frac{1}{c^{2}}$ and $q=p$ then we have $A||u||_{L^{p}} leq ||Du||_{L^{p}}$ then since $||u||_{W_{0}^{1,p}} leq A||u||_{L^{p}} + A||Du||_{L^{p}} leq (1+A)||Du||_{L^{p}}$ it follows that $frac{A}{A+1}||u||_{W_{0}^{1,p}} leq ||Du||_{L^{p}}$
Take $B_{1} := frac{A}{A+1}$.
The result follows from combining the two inequalities.
answered Oct 22 '13 at 11:33
AlexAlex
189218
add a comment |
$$
|Du|_{L^p(U)} le |Du|_{L^p(U)}+|u|_{L^p(U)} = |u|_{W_0^{1,p}(U)} le (1+C),|Du|_{L^p(U)}
$$
where $C$ is the constant in Poincare's inequality.
edited Nov 30 '18 at 1:51
bosmacs
547316
answered Oct 22 '13 at 10:18
Julián AguirreJulián Aguirre
67.9k24094
1
Dear @Julián, I think it is good to correct the typo in your answer: $|Du|_{L^p(U)}+|u|_{L^p(U)}=|u|_{W_0^{1,p}(U)}$ instead of $|Du|_{L^p(U)}+|u|_{L^p(U)}=|Du|_{W_0^{1,p}(U)}$. This may cause some confusion in the beginners.
– Tomás
Oct 23 '13 at 19:24
@Tomás Thank you for your comment. I have edited my answer.
– Julián Aguirre
Oct 24 '13 at 15:17
Dear @Julián, sorry to bother you again, but you have changed to wrong place.
– Tomás
Oct 24 '13 at 15:32
@Tomás I should not work at "siesta" time. Thanks again.
– Julián Aguirre
Oct 24 '13 at 15:36
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: "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%2f535503%2fpoincare-inequality-implies-equivalent-norms%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
This applies to $1 leq p < n$. You want constants $B_{1}$ and $B_{2}$ such that $B_{1}||u||_{W_{0}^{1,p}(U)} leq ||Du||_{L^{p}(U)} leq B_{2}||u||_{W_{0}^{1,p}}$. The inequality $||Du||_{L^{p}(U)} leq B_{2}||u||_{W_{0}^{1,p}}$ is trivial.
From Gagliardo-Nirenberg-Sobolev Inequality we have:
$||u||_{L^{p*}(U)} leq C||Du||_{L^{p}(U)}$ where $p* := frac{np}{n-p}$ $p* > p$.
We also have $||u||_{L^{q}(U)} leq C||u||_{L^{p*}(U)}$ if $1leq q leq p*$ by Generalized Holder Inequality.
If we take $A = frac{1}{c^{2}}$ and $q=p$ then we have $A||u||_{L^{p}} leq ||Du||_{L^{p}}$ then since $||u||_{W_{0}^{1,p}} leq A||u||_{L^{p}} + A||Du||_{L^{p}} leq (1+A)||Du||_{L^{p}}$ it follows that $frac{A}{A+1}||u||_{W_{0}^{1,p}} leq ||Du||_{L^{p}}$
Take $B_{1} := frac{A}{A+1}$.
The result follows from combining the two inequalities.
answered Oct 22 '13 at 11:33
AlexAlex
189218
add a comment |
This applies to $1 leq p < n$. You want constants $B_{1}$ and $B_{2}$ such that $B_{1}||u||_{W_{0}^{1,p}(U)} leq ||Du||_{L^{p}(U)} leq B_{2}||u||_{W_{0}^{1,p}}$. The inequality $||Du||_{L^{p}(U)} leq B_{2}||u||_{W_{0}^{1,p}}$ is trivial.
From Gagliardo-Nirenberg-Sobolev Inequality we have:
$||u||_{L^{p*}(U)} leq C||Du||_{L^{p}(U)}$ where $p* := frac{np}{n-p}$ $p* > p$.
We also have $||u||_{L^{q}(U)} leq C||u||_{L^{p*}(U)}$ if $1leq q leq p*$ by Generalized Holder Inequality.
If we take $A = frac{1}{c^{2}}$ and $q=p$ then we have $A||u||_{L^{p}} leq ||Du||_{L^{p}}$ then since $||u||_{W_{0}^{1,p}} leq A||u||_{L^{p}} + A||Du||_{L^{p}} leq (1+A)||Du||_{L^{p}}$ it follows that $frac{A}{A+1}||u||_{W_{0}^{1,p}} leq ||Du||_{L^{p}}$
Take $B_{1} := frac{A}{A+1}$.
The result follows from combining the two inequalities.
answered Oct 22 '13 at 11:33
AlexAlex
189218
add a comment |
This applies to $1 leq p < n$. You want constants $B_{1}$ and $B_{2}$ such that $B_{1}||u||_{W_{0}^{1,p}(U)} leq ||Du||_{L^{p}(U)} leq B_{2}||u||_{W_{0}^{1,p}}$. The inequality $||Du||_{L^{p}(U)} leq B_{2}||u||_{W_{0}^{1,p}}$ is trivial.
From Gagliardo-Nirenberg-Sobolev Inequality we have:
$||u||_{L^{p*}(U)} leq C||Du||_{L^{p}(U)}$ where $p* := frac{np}{n-p}$ $p* > p$.
We also have $||u||_{L^{q}(U)} leq C||u||_{L^{p*}(U)}$ if $1leq q leq p*$ by Generalized Holder Inequality.
If we take $A = frac{1}{c^{2}}$ and $q=p$ then we have $A||u||_{L^{p}} leq ||Du||_{L^{p}}$ then since $||u||_{W_{0}^{1,p}} leq A||u||_{L^{p}} + A||Du||_{L^{p}} leq (1+A)||Du||_{L^{p}}$ it follows that $frac{A}{A+1}||u||_{W_{0}^{1,p}} leq ||Du||_{L^{p}}$
Take $B_{1} := frac{A}{A+1}$.
The result follows from combining the two inequalities.
answered Oct 22 '13 at 11:33
AlexAlex
189218
This applies to $1 leq p < n$. You want constants $B_{1}$ and $B_{2}$ such that $B_{1}||u||_{W_{0}^{1,p}(U)} leq ||Du||_{L^{p}(U)} leq B_{2}||u||_{W_{0}^{1,p}}$. The inequality $||Du||_{L^{p}(U)} leq B_{2}||u||_{W_{0}^{1,p}}$ is trivial.
From Gagliardo-Nirenberg-Sobolev Inequality we have:
$||u||_{L^{p*}(U)} leq C||Du||_{L^{p}(U)}$ where $p* := frac{np}{n-p}$ $p* > p$.
We also have $||u||_{L^{q}(U)} leq C||u||_{L^{p*}(U)}$ if $1leq q leq p*$ by Generalized Holder Inequality.
If we take $A = frac{1}{c^{2}}$ and $q=p$ then we have $A||u||_{L^{p}} leq ||Du||_{L^{p}}$ then since $||u||_{W_{0}^{1,p}} leq A||u||_{L^{p}} + A||Du||_{L^{p}} leq (1+A)||Du||_{L^{p}}$ it follows that $frac{A}{A+1}||u||_{W_{0}^{1,p}} leq ||Du||_{L^{p}}$
Take $B_{1} := frac{A}{A+1}$.
The result follows from combining the two inequalities.
answered Oct 22 '13 at 11:33
AlexAlex
189218
answered Oct 22 '13 at 11:33
AlexAlex
189218
answered Oct 22 '13 at 11:33
AlexAlex
189218
answered Oct 22 '13 at 11:33
AlexAlex
189218
189218
add a comment |
add a comment |
$$
|Du|_{L^p(U)} le |Du|_{L^p(U)}+|u|_{L^p(U)} = |u|_{W_0^{1,p}(U)} le (1+C),|Du|_{L^p(U)}
$$
where $C$ is the constant in Poincare's inequality.
edited Nov 30 '18 at 1:51
bosmacs
547316
answered Oct 22 '13 at 10:18
Julián AguirreJulián Aguirre
67.9k24094
1
Dear @Julián, I think it is good to correct the typo in your answer: $|Du|_{L^p(U)}+|u|_{L^p(U)}=|u|_{W_0^{1,p}(U)}$ instead of $|Du|_{L^p(U)}+|u|_{L^p(U)}=|Du|_{W_0^{1,p}(U)}$. This may cause some confusion in the beginners.
– Tomás
Oct 23 '13 at 19:24
@Tomás Thank you for your comment. I have edited my answer.
– Julián Aguirre
Oct 24 '13 at 15:17
Dear @Julián, sorry to bother you again, but you have changed to wrong place.
– Tomás
Oct 24 '13 at 15:32
@Tomás I should not work at "siesta" time. Thanks again.
– Julián Aguirre
Oct 24 '13 at 15:36
add a comment |
$$
|Du|_{L^p(U)} le |Du|_{L^p(U)}+|u|_{L^p(U)} = |u|_{W_0^{1,p}(U)} le (1+C),|Du|_{L^p(U)}
$$
where $C$ is the constant in Poincare's inequality.
edited Nov 30 '18 at 1:51
bosmacs
547316
answered Oct 22 '13 at 10:18
Julián AguirreJulián Aguirre
67.9k24094
1
Dear @Julián, I think it is good to correct the typo in your answer: $|Du|_{L^p(U)}+|u|_{L^p(U)}=|u|_{W_0^{1,p}(U)}$ instead of $|Du|_{L^p(U)}+|u|_{L^p(U)}=|Du|_{W_0^{1,p}(U)}$. This may cause some confusion in the beginners.
– Tomás
Oct 23 '13 at 19:24
@Tomás Thank you for your comment. I have edited my answer.
– Julián Aguirre
Oct 24 '13 at 15:17
Dear @Julián, sorry to bother you again, but you have changed to wrong place.
– Tomás
Oct 24 '13 at 15:32
@Tomás I should not work at "siesta" time. Thanks again.
– Julián Aguirre
Oct 24 '13 at 15:36
add a comment |
$$
|Du|_{L^p(U)} le |Du|_{L^p(U)}+|u|_{L^p(U)} = |u|_{W_0^{1,p}(U)} le (1+C),|Du|_{L^p(U)}
$$
where $C$ is the constant in Poincare's inequality.
edited Nov 30 '18 at 1:51
bosmacs
547316
answered Oct 22 '13 at 10:18
Julián AguirreJulián Aguirre
67.9k24094
$$
|Du|_{L^p(U)} le |Du|_{L^p(U)}+|u|_{L^p(U)} = |u|_{W_0^{1,p}(U)} le (1+C),|Du|_{L^p(U)}
$$
where $C$ is the constant in Poincare's inequality.
edited Nov 30 '18 at 1:51
bosmacs
547316
answered Oct 22 '13 at 10:18
Julián AguirreJulián Aguirre
67.9k24094
edited Nov 30 '18 at 1:51
bosmacs
547316
edited Nov 30 '18 at 1:51
bosmacs
547316
edited Nov 30 '18 at 1:51
bosmacs
547316
547316
answered Oct 22 '13 at 10:18
Julián AguirreJulián Aguirre
67.9k24094
answered Oct 22 '13 at 10:18
Julián AguirreJulián Aguirre
67.9k24094
answered Oct 22 '13 at 10:18
Julián AguirreJulián Aguirre
67.9k24094
67.9k24094
1
Dear @Julián, I think it is good to correct the typo in your answer: $|Du|_{L^p(U)}+|u|_{L^p(U)}=|u|_{W_0^{1,p}(U)}$ instead of $|Du|_{L^p(U)}+|u|_{L^p(U)}=|Du|_{W_0^{1,p}(U)}$. This may cause some confusion in the beginners.
– Tomás
Oct 23 '13 at 19:24
@Tomás Thank you for your comment. I have edited my answer.
– Julián Aguirre
Oct 24 '13 at 15:17
Dear @Julián, sorry to bother you again, but you have changed to wrong place.
– Tomás
Oct 24 '13 at 15:32
@Tomás I should not work at "siesta" time. Thanks again.
– Julián Aguirre
Oct 24 '13 at 15:36
add a comment |
1
Dear @Julián, I think it is good to correct the typo in your answer: $|Du|_{L^p(U)}+|u|_{L^p(U)}=|u|_{W_0^{1,p}(U)}$ instead of $|Du|_{L^p(U)}+|u|_{L^p(U)}=|Du|_{W_0^{1,p}(U)}$. This may cause some confusion in the beginners.
– Tomás
Oct 23 '13 at 19:24
@Tomás Thank you for your comment. I have edited my answer.
– Julián Aguirre
Oct 24 '13 at 15:17
Dear @Julián, sorry to bother you again, but you have changed to wrong place.
– Tomás
Oct 24 '13 at 15:32
@Tomás I should not work at "siesta" time. Thanks again.
– Julián Aguirre
Oct 24 '13 at 15:36
1
1
Dear @Julián, I think it is good to correct the typo in your answer: $|Du|_{L^p(U)}+|u|_{L^p(U)}=|u|_{W_0^{1,p}(U)}$ instead of $|Du|_{L^p(U)}+|u|_{L^p(U)}=|Du|_{W_0^{1,p}(U)}$. This may cause some confusion in the beginners.
– Tomás
Oct 23 '13 at 19:24
Dear @Julián, I think it is good to correct the typo in your answer: $|Du|_{L^p(U)}+|u|_{L^p(U)}=|u|_{W_0^{1,p}(U)}$ instead of $|Du|_{L^p(U)}+|u|_{L^p(U)}=|Du|_{W_0^{1,p}(U)}$. This may cause some confusion in the beginners.
– Tomás
Oct 23 '13 at 19:24
@Tomás Thank you for your comment. I have edited my answer.
– Julián Aguirre
Oct 24 '13 at 15:17
@Tomás Thank you for your comment. I have edited my answer.
– Julián Aguirre
Oct 24 '13 at 15:17
Dear @Julián, sorry to bother you again, but you have changed to wrong place.
– Tomás
Oct 24 '13 at 15:32
Dear @Julián, sorry to bother you again, but you have changed to wrong place.
– Tomás
Oct 24 '13 at 15:32
@Tomás I should not work at "siesta" time. Thanks again.
– Julián Aguirre
Oct 24 '13 at 15:36
@Tomás I should not work at "siesta" time. Thanks again.
– Julián Aguirre
Oct 24 '13 at 15:36
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.
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%2f535503%2fpoincare-inequality-implies-equivalent-norms%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
I can't understand why do you accepted Alex answers if he gives you only a partial answer (also to complicated).
– Tomás
Oct 23 '13 at 18:30
I don't understand why you think it is a partial answer? It seems fine to me. He is doing it using results from Evans book.
– Lucio D
Oct 23 '13 at 18:38
He considered only the case $1leq p< n$, while Julian proof is valid for all cases.
– Tomás
Oct 23 '13 at 19:02
Yes I agree Julien's is simpler and stronger answer seeing as it is more general. The Remark in Evans is referring to that particular case of $1 leq p < n$. I just ticked Alex's because he used the proofs presented in Evans up until that point, so it followed well in the sequence in which I am studying it. But I indicated 'useful answer for both'. I think Julien made a type though, it should be $||Du||_{L^{p}(U)} leq ||Du||_{L^{p}(U)} + ||u||_{L^{p}(U)} = ||u||_{W_{0}^{1,p}(U)} leq ||Du||_{L^{p}(U)} + C||Du||_{L^{p}(U)}$.
– Lucio D
Oct 23 '13 at 19:16
Ok, in the end you are aware that the inequality is valid for all $p$.
– Tomás
Oct 23 '13 at 19:19