Mixing
Does functional satisfies Palais-Smale condition?
$begingroup$
Check if the functional $f(u)=int_0^{1/2} u^2(x)dx$ satisfies the Palais-Smale condition on the Hilbert space $L^2([0,1],mathbb{R})$.
We have definied the Palais-Smale condition as follows:
$f$ satisfies the PS-condition if every sequence $(u_k)subset L^2([0,1],mathbb{R})$ which satisfies $f(u_k)$ is bounded and $Df(u_k) rightarrow 0$ in $L^2([0,1],mathbb{R})$, there exists a convergent subsequence.
I think that this functional satisfies the Palais-Smale condition (is this true?) but I don't really know how to prove this correctly.
My idea was to take a PS-sequence $(u_k)$, that means $f(u_k)$ is bounded and $Df(u_k) = 2 int_0^{1/2} u_k(x) nabla u_k(x) dx rightarrow 0$ in $L^2([0,1],mathbb{R})$. And I thought that maybe to prove this, I could show that this PS-sequence is bounded. Then I could follow with Banach-Alaoglu that $u_k$ has a subsequence which converges weakly. Is this idea reasonable or can't I prove the assumption like that?
I wanted to show that $|u_k|^2$ is bounded (with the $L^2$-norm) which implies that $u_k$ is bounded and is easier to show, because we know that $|u_k|^2 = (u_k,u_k) = int_0^1 u_k^2(x)dx$. But even if I know that $f(u_k)$ is bounded, I wasn't able to show that the integral is bounded on the whole interval $[0,1]$. Maybe someone could give me some tips to solve this problem.
real-analysis calculus-of-variations variational-analysis
asked Jan 6 at 18:09
mathstumathstu
314
$endgroup$
add a comment |
$begingroup$
Check if the functional $f(u)=int_0^{1/2} u^2(x)dx$ satisfies the Palais-Smale condition on the Hilbert space $L^2([0,1],mathbb{R})$.
We have definied the Palais-Smale condition as follows:
$f$ satisfies the PS-condition if every sequence $(u_k)subset L^2([0,1],mathbb{R})$ which satisfies $f(u_k)$ is bounded and $Df(u_k) rightarrow 0$ in $L^2([0,1],mathbb{R})$, there exists a convergent subsequence.
I think that this functional satisfies the Palais-Smale condition (is this true?) but I don't really know how to prove this correctly.
My idea was to take a PS-sequence $(u_k)$, that means $f(u_k)$ is bounded and $Df(u_k) = 2 int_0^{1/2} u_k(x) nabla u_k(x) dx rightarrow 0$ in $L^2([0,1],mathbb{R})$. And I thought that maybe to prove this, I could show that this PS-sequence is bounded. Then I could follow with Banach-Alaoglu that $u_k$ has a subsequence which converges weakly. Is this idea reasonable or can't I prove the assumption like that?
I wanted to show that $|u_k|^2$ is bounded (with the $L^2$-norm) which implies that $u_k$ is bounded and is easier to show, because we know that $|u_k|^2 = (u_k,u_k) = int_0^1 u_k^2(x)dx$. But even if I know that $f(u_k)$ is bounded, I wasn't able to show that the integral is bounded on the whole interval $[0,1]$. Maybe someone could give me some tips to solve this problem.
real-analysis calculus-of-variations variational-analysis
asked Jan 6 at 18:09
mathstumathstu
314
$endgroup$
2
$begingroup$
I doubt that it satisfies the condition on $L^2(0,1)$. The functional $f$ only detects the behavior of the sequence $left{u_kright}$ in $(0,1/2)$. So, on $(1/2,1)$ it can do whatever it wants, but to satisfy the $PS$ condition $left{u_kright}$ needs to have a converging subsequence on $L^2(0,1)$.
$endgroup$
– Lorenzo Quarisa
Jan 6 at 22:27
add a comment |
$begingroup$
Check if the functional $f(u)=int_0^{1/2} u^2(x)dx$ satisfies the Palais-Smale condition on the Hilbert space $L^2([0,1],mathbb{R})$.
We have definied the Palais-Smale condition as follows:
$f$ satisfies the PS-condition if every sequence $(u_k)subset L^2([0,1],mathbb{R})$ which satisfies $f(u_k)$ is bounded and $Df(u_k) rightarrow 0$ in $L^2([0,1],mathbb{R})$, there exists a convergent subsequence.
I think that this functional satisfies the Palais-Smale condition (is this true?) but I don't really know how to prove this correctly.
My idea was to take a PS-sequence $(u_k)$, that means $f(u_k)$ is bounded and $Df(u_k) = 2 int_0^{1/2} u_k(x) nabla u_k(x) dx rightarrow 0$ in $L^2([0,1],mathbb{R})$. And I thought that maybe to prove this, I could show that this PS-sequence is bounded. Then I could follow with Banach-Alaoglu that $u_k$ has a subsequence which converges weakly. Is this idea reasonable or can't I prove the assumption like that?
I wanted to show that $|u_k|^2$ is bounded (with the $L^2$-norm) which implies that $u_k$ is bounded and is easier to show, because we know that $|u_k|^2 = (u_k,u_k) = int_0^1 u_k^2(x)dx$. But even if I know that $f(u_k)$ is bounded, I wasn't able to show that the integral is bounded on the whole interval $[0,1]$. Maybe someone could give me some tips to solve this problem.
real-analysis calculus-of-variations variational-analysis
asked Jan 6 at 18:09
mathstumathstu
314
$endgroup$
Check if the functional $f(u)=int_0^{1/2} u^2(x)dx$ satisfies the Palais-Smale condition on the Hilbert space $L^2([0,1],mathbb{R})$.
We have definied the Palais-Smale condition as follows:
$f$ satisfies the PS-condition if every sequence $(u_k)subset L^2([0,1],mathbb{R})$ which satisfies $f(u_k)$ is bounded and $Df(u_k) rightarrow 0$ in $L^2([0,1],mathbb{R})$, there exists a convergent subsequence.
I think that this functional satisfies the Palais-Smale condition (is this true?) but I don't really know how to prove this correctly.
My idea was to take a PS-sequence $(u_k)$, that means $f(u_k)$ is bounded and $Df(u_k) = 2 int_0^{1/2} u_k(x) nabla u_k(x) dx rightarrow 0$ in $L^2([0,1],mathbb{R})$. And I thought that maybe to prove this, I could show that this PS-sequence is bounded. Then I could follow with Banach-Alaoglu that $u_k$ has a subsequence which converges weakly. Is this idea reasonable or can't I prove the assumption like that?
I wanted to show that $|u_k|^2$ is bounded (with the $L^2$-norm) which implies that $u_k$ is bounded and is easier to show, because we know that $|u_k|^2 = (u_k,u_k) = int_0^1 u_k^2(x)dx$. But even if I know that $f(u_k)$ is bounded, I wasn't able to show that the integral is bounded on the whole interval $[0,1]$. Maybe someone could give me some tips to solve this problem.
real-analysis calculus-of-variations variational-analysis
real-analysis calculus-of-variations variational-analysis
asked Jan 6 at 18:09
mathstumathstu
314
asked Jan 6 at 18:09
mathstumathstu
314
asked Jan 6 at 18:09
mathstumathstu
314
asked Jan 6 at 18:09
mathstumathstu
314
asked Jan 6 at 18:09
mathstumathstu
314
314
2
$begingroup$
I doubt that it satisfies the condition on $L^2(0,1)$. The functional $f$ only detects the behavior of the sequence $left{u_kright}$ in $(0,1/2)$. So, on $(1/2,1)$ it can do whatever it wants, but to satisfy the $PS$ condition $left{u_kright}$ needs to have a converging subsequence on $L^2(0,1)$.
$endgroup$
– Lorenzo Quarisa
Jan 6 at 22:27
add a comment |
2
$begingroup$
I doubt that it satisfies the condition on $L^2(0,1)$. The functional $f$ only detects the behavior of the sequence $left{u_kright}$ in $(0,1/2)$. So, on $(1/2,1)$ it can do whatever it wants, but to satisfy the $PS$ condition $left{u_kright}$ needs to have a converging subsequence on $L^2(0,1)$.
$endgroup$
– Lorenzo Quarisa
Jan 6 at 22:27
2
2
$begingroup$
I doubt that it satisfies the condition on $L^2(0,1)$. The functional $f$ only detects the behavior of the sequence $left{u_kright}$ in $(0,1/2)$. So, on $(1/2,1)$ it can do whatever it wants, but to satisfy the $PS$ condition $left{u_kright}$ needs to have a converging subsequence on $L^2(0,1)$.
$endgroup$
– Lorenzo Quarisa
Jan 6 at 22:27
$begingroup$
I doubt that it satisfies the condition on $L^2(0,1)$. The functional $f$ only detects the behavior of the sequence $left{u_kright}$ in $(0,1/2)$. So, on $(1/2,1)$ it can do whatever it wants, but to satisfy the $PS$ condition $left{u_kright}$ needs to have a converging subsequence on $L^2(0,1)$.
$endgroup$
– Lorenzo Quarisa
Jan 6 at 22:27
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%2f3064207%2fdoes-functional-satisfies-palais-smale-condition%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%2f3064207%2fdoes-functional-satisfies-palais-smale-condition%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
2
$begingroup$
I doubt that it satisfies the condition on $L^2(0,1)$. The functional $f$ only detects the behavior of the sequence $left{u_kright}$ in $(0,1/2)$. So, on $(1/2,1)$ it can do whatever it wants, but to satisfy the $PS$ condition $left{u_kright}$ needs to have a converging subsequence on $L^2(0,1)$.
$endgroup$
– Lorenzo Quarisa
Jan 6 at 22:27