Periodic boundary solution forms a closed subspace in $H^1$
I am reading the Sobolev spaces chapter in Functional Analysis by Brezis. And there is a statement saying that
$H = left{v in H^1(0,1)| v(0) = v(1) right}$ is a closed subspace of $H^1(0,1)$. I am wondering why $v_n rightarrow v$ in $H^1(0,1)$ and $v_n(0) = v_n(1)$ can conclude that $v(0) = v(1)$.
Any help will be appreciated!
functional-analysis
asked Dec 2 at 16:31
lxnlll
344
add a comment |
I am reading the Sobolev spaces chapter in Functional Analysis by Brezis. And there is a statement saying that
$H = left{v in H^1(0,1)| v(0) = v(1) right}$ is a closed subspace of $H^1(0,1)$. I am wondering why $v_n rightarrow v$ in $H^1(0,1)$ and $v_n(0) = v_n(1)$ can conclude that $v(0) = v(1)$.
Any help will be appreciated!
functional-analysis
asked Dec 2 at 16:31
lxnlll
344
add a comment |
I am reading the Sobolev spaces chapter in Functional Analysis by Brezis. And there is a statement saying that
$H = left{v in H^1(0,1)| v(0) = v(1) right}$ is a closed subspace of $H^1(0,1)$. I am wondering why $v_n rightarrow v$ in $H^1(0,1)$ and $v_n(0) = v_n(1)$ can conclude that $v(0) = v(1)$.
Any help will be appreciated!
functional-analysis
asked Dec 2 at 16:31
lxnlll
344
I am reading the Sobolev spaces chapter in Functional Analysis by Brezis. And there is a statement saying that
$H = left{v in H^1(0,1)| v(0) = v(1) right}$ is a closed subspace of $H^1(0,1)$. I am wondering why $v_n rightarrow v$ in $H^1(0,1)$ and $v_n(0) = v_n(1)$ can conclude that $v(0) = v(1)$.
Any help will be appreciated!
functional-analysis
functional-analysis
asked Dec 2 at 16:31
lxnlll
344
asked Dec 2 at 16:31
lxnlll
344
asked Dec 2 at 16:31
lxnlll
344
asked Dec 2 at 16:31
lxnlll
344
asked Dec 2 at 16:31
lxnlll
344
344
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
The functional
$$
u mapsto u(1) - u(0)
$$
is linear and bounded on $H^1(0,1)$. Hence, its kernel is closed and this kernel coincides with the subspace of periodic functions.
answered Dec 3 at 7:27
gerw
19k11133
add a comment |
Maybe this works:
$$v(0) = v(0) - v(1) + v(1) = -int_0^1 v'dt + v(1) = lim_{n to infty} int_0^1 v_n' dt + v(1) = v(1),$$
where $v_n', v'$ denote the weak derivatives of $v_n, v$.
answered Dec 2 at 17:52
eddie
325110
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%2f3022845%2fperiodic-boundary-solution-forms-a-closed-subspace-in-h1%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
The functional
$$
u mapsto u(1) - u(0)
$$
is linear and bounded on $H^1(0,1)$. Hence, its kernel is closed and this kernel coincides with the subspace of periodic functions.
answered Dec 3 at 7:27
gerw
19k11133
add a comment |
The functional
$$
u mapsto u(1) - u(0)
$$
is linear and bounded on $H^1(0,1)$. Hence, its kernel is closed and this kernel coincides with the subspace of periodic functions.
answered Dec 3 at 7:27
gerw
19k11133
add a comment |
The functional
$$
u mapsto u(1) - u(0)
$$
is linear and bounded on $H^1(0,1)$. Hence, its kernel is closed and this kernel coincides with the subspace of periodic functions.
answered Dec 3 at 7:27
gerw
19k11133
The functional
$$
u mapsto u(1) - u(0)
$$
is linear and bounded on $H^1(0,1)$. Hence, its kernel is closed and this kernel coincides with the subspace of periodic functions.
answered Dec 3 at 7:27
gerw
19k11133
answered Dec 3 at 7:27
gerw
19k11133
answered Dec 3 at 7:27
gerw
19k11133
answered Dec 3 at 7:27
gerw
19k11133
19k11133
add a comment |
add a comment |
Maybe this works:
$$v(0) = v(0) - v(1) + v(1) = -int_0^1 v'dt + v(1) = lim_{n to infty} int_0^1 v_n' dt + v(1) = v(1),$$
where $v_n', v'$ denote the weak derivatives of $v_n, v$.
answered Dec 2 at 17:52
eddie
325110
add a comment |
Maybe this works:
$$v(0) = v(0) - v(1) + v(1) = -int_0^1 v'dt + v(1) = lim_{n to infty} int_0^1 v_n' dt + v(1) = v(1),$$
where $v_n', v'$ denote the weak derivatives of $v_n, v$.
answered Dec 2 at 17:52
eddie
325110
add a comment |
Maybe this works:
$$v(0) = v(0) - v(1) + v(1) = -int_0^1 v'dt + v(1) = lim_{n to infty} int_0^1 v_n' dt + v(1) = v(1),$$
where $v_n', v'$ denote the weak derivatives of $v_n, v$.
answered Dec 2 at 17:52
eddie
325110
Maybe this works:
$$v(0) = v(0) - v(1) + v(1) = -int_0^1 v'dt + v(1) = lim_{n to infty} int_0^1 v_n' dt + v(1) = v(1),$$
where $v_n', v'$ denote the weak derivatives of $v_n, v$.
answered Dec 2 at 17:52
eddie
325110
answered Dec 2 at 17:52
eddie
325110
answered Dec 2 at 17:52
eddie
325110
answered Dec 2 at 17:52
eddie
325110
325110
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3022845%2fperiodic-boundary-solution-forms-a-closed-subspace-in-h1%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
