What is the orthogonal complement of $H^1_0$ in $H^1$?












6














Let $Omega$ be a closed domain with smooth boundary in $mathbb{R}^n$. Let $H^1_0(Omega)$ be the closure of compactly supported smooth functions under the norm $|u|_1 = int_Omega u^2 + |nabla u|^2 dx$ and let $H^1(Omega)$ be the closure of smooth, continuous functions under the same norm.



Any $H^1$ function which has nonvanishing trace cannot be approximated by any sequence of functions in $H^1_0$. So $H^1_0$ is a closed subspace of the Hilbert space $(H^1, |cdot|_1)$, hence has an orthogonal complement.




What is a generating set of the orthogonal complement of $H^1_0$ in $H^1$?




Motivation is to get my hands on some concrete examples, rather than to just appeal to theorems that establish the existence of a right inverse to a trace operator.



Of course if anyone has references, I'm happy to follow them up. I've skimmed through Gilbarg-Trudinger and Evans and found nothing, but maybe I'm looking in the wrong place.






functional-analysis sobolev-spaces trace







share|cite|improve this question






















  • For $H^1([0,1])$, say $langle f,g rangle_{H^1([0,1])} = f(0)overline{g(0)}+f(1)overline{g(1)}+langle f',g' rangle_{L^2([0,1])}$. $H^1_0([0,1]) = { f in H^1([0,1]),f(0) = f(1)=0}={ f in H^1([0,1]),f(0) = f(1)=langle f',1 rangle_{L^2([0,1])}=0}$ so the orthogonal complement is ${a+bx}$. When changing the inner product (for say $(f,g ) = langle f,g rangle_{L^2([0,1])}+langle f',g' rangle_{L^2([0,1])}$) the orthogonal complement becomes ${a phi_1+b phi_2}$
    – reuns
    Nov 20 at 1:54












  • @reuns surely there is a condition on $a,b$? With $f(x) = begin{cases}x, & x < frac{1}{2}\ 1-x, & x geq frac{1}{2}end{cases}$ and $g(x) = a + bx$ I compute $langle f, grangle_{H^1([0,1])} = frac{1}{4}a + frac{1}{8}b$ which is not always zero
    – Neal
    Nov 20 at 2:04












  • That's why I gave two different inner products, only for the first one the orthogonal complement has an obvious basis, for the other inner product, I'd say we should substract to $1$ and $x$ their projection on the trigonometric orthogonal basis of $H^1_0$
    – reuns
    Nov 20 at 2:13










  • @reuns Ah, I see, thank you. Your comments make good sense. (I would note that I specified the inner product in the question.)
    – Neal
    Nov 20 at 13:33


















6














Let $Omega$ be a closed domain with smooth boundary in $mathbb{R}^n$. Let $H^1_0(Omega)$ be the closure of compactly supported smooth functions under the norm $|u|_1 = int_Omega u^2 + |nabla u|^2 dx$ and let $H^1(Omega)$ be the closure of smooth, continuous functions under the same norm.



Any $H^1$ function which has nonvanishing trace cannot be approximated by any sequence of functions in $H^1_0$. So $H^1_0$ is a closed subspace of the Hilbert space $(H^1, |cdot|_1)$, hence has an orthogonal complement.




What is a generating set of the orthogonal complement of $H^1_0$ in $H^1$?




Motivation is to get my hands on some concrete examples, rather than to just appeal to theorems that establish the existence of a right inverse to a trace operator.



Of course if anyone has references, I'm happy to follow them up. I've skimmed through Gilbarg-Trudinger and Evans and found nothing, but maybe I'm looking in the wrong place.






functional-analysis sobolev-spaces trace







share|cite|improve this question






















  • For $H^1([0,1])$, say $langle f,g rangle_{H^1([0,1])} = f(0)overline{g(0)}+f(1)overline{g(1)}+langle f',g' rangle_{L^2([0,1])}$. $H^1_0([0,1]) = { f in H^1([0,1]),f(0) = f(1)=0}={ f in H^1([0,1]),f(0) = f(1)=langle f',1 rangle_{L^2([0,1])}=0}$ so the orthogonal complement is ${a+bx}$. When changing the inner product (for say $(f,g ) = langle f,g rangle_{L^2([0,1])}+langle f',g' rangle_{L^2([0,1])}$) the orthogonal complement becomes ${a phi_1+b phi_2}$
    – reuns
    Nov 20 at 1:54












  • @reuns surely there is a condition on $a,b$? With $f(x) = begin{cases}x, & x < frac{1}{2}\ 1-x, & x geq frac{1}{2}end{cases}$ and $g(x) = a + bx$ I compute $langle f, grangle_{H^1([0,1])} = frac{1}{4}a + frac{1}{8}b$ which is not always zero
    – Neal
    Nov 20 at 2:04












  • That's why I gave two different inner products, only for the first one the orthogonal complement has an obvious basis, for the other inner product, I'd say we should substract to $1$ and $x$ their projection on the trigonometric orthogonal basis of $H^1_0$
    – reuns
    Nov 20 at 2:13










  • @reuns Ah, I see, thank you. Your comments make good sense. (I would note that I specified the inner product in the question.)
    – Neal
    Nov 20 at 13:33
















6












6








6


1





Let $Omega$ be a closed domain with smooth boundary in $mathbb{R}^n$. Let $H^1_0(Omega)$ be the closure of compactly supported smooth functions under the norm $|u|_1 = int_Omega u^2 + |nabla u|^2 dx$ and let $H^1(Omega)$ be the closure of smooth, continuous functions under the same norm.



Any $H^1$ function which has nonvanishing trace cannot be approximated by any sequence of functions in $H^1_0$. So $H^1_0$ is a closed subspace of the Hilbert space $(H^1, |cdot|_1)$, hence has an orthogonal complement.




What is a generating set of the orthogonal complement of $H^1_0$ in $H^1$?




Motivation is to get my hands on some concrete examples, rather than to just appeal to theorems that establish the existence of a right inverse to a trace operator.



Of course if anyone has references, I'm happy to follow them up. I've skimmed through Gilbarg-Trudinger and Evans and found nothing, but maybe I'm looking in the wrong place.






functional-analysis sobolev-spaces trace







share|cite|improve this question













Let $Omega$ be a closed domain with smooth boundary in $mathbb{R}^n$. Let $H^1_0(Omega)$ be the closure of compactly supported smooth functions under the norm $|u|_1 = int_Omega u^2 + |nabla u|^2 dx$ and let $H^1(Omega)$ be the closure of smooth, continuous functions under the same norm.



Any $H^1$ function which has nonvanishing trace cannot be approximated by any sequence of functions in $H^1_0$. So $H^1_0$ is a closed subspace of the Hilbert space $(H^1, |cdot|_1)$, hence has an orthogonal complement.




What is a generating set of the orthogonal complement of $H^1_0$ in $H^1$?




Motivation is to get my hands on some concrete examples, rather than to just appeal to theorems that establish the existence of a right inverse to a trace operator.



Of course if anyone has references, I'm happy to follow them up. I've skimmed through Gilbarg-Trudinger and Evans and found nothing, but maybe I'm looking in the wrong place.






functional-analysis sobolev-spaces trace




functional-analysis sobolev-spaces trace






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 20 at 1:34









Neal

23.5k23683




23.5k23683












  • For $H^1([0,1])$, say $langle f,g rangle_{H^1([0,1])} = f(0)overline{g(0)}+f(1)overline{g(1)}+langle f',g' rangle_{L^2([0,1])}$. $H^1_0([0,1]) = { f in H^1([0,1]),f(0) = f(1)=0}={ f in H^1([0,1]),f(0) = f(1)=langle f',1 rangle_{L^2([0,1])}=0}$ so the orthogonal complement is ${a+bx}$. When changing the inner product (for say $(f,g ) = langle f,g rangle_{L^2([0,1])}+langle f',g' rangle_{L^2([0,1])}$) the orthogonal complement becomes ${a phi_1+b phi_2}$
    – reuns
    Nov 20 at 1:54












  • @reuns surely there is a condition on $a,b$? With $f(x) = begin{cases}x, & x < frac{1}{2}\ 1-x, & x geq frac{1}{2}end{cases}$ and $g(x) = a + bx$ I compute $langle f, grangle_{H^1([0,1])} = frac{1}{4}a + frac{1}{8}b$ which is not always zero
    – Neal
    Nov 20 at 2:04












  • That's why I gave two different inner products, only for the first one the orthogonal complement has an obvious basis, for the other inner product, I'd say we should substract to $1$ and $x$ their projection on the trigonometric orthogonal basis of $H^1_0$
    – reuns
    Nov 20 at 2:13










  • @reuns Ah, I see, thank you. Your comments make good sense. (I would note that I specified the inner product in the question.)
    – Neal
    Nov 20 at 13:33




















  • For $H^1([0,1])$, say $langle f,g rangle_{H^1([0,1])} = f(0)overline{g(0)}+f(1)overline{g(1)}+langle f',g' rangle_{L^2([0,1])}$. $H^1_0([0,1]) = { f in H^1([0,1]),f(0) = f(1)=0}={ f in H^1([0,1]),f(0) = f(1)=langle f',1 rangle_{L^2([0,1])}=0}$ so the orthogonal complement is ${a+bx}$. When changing the inner product (for say $(f,g ) = langle f,g rangle_{L^2([0,1])}+langle f',g' rangle_{L^2([0,1])}$) the orthogonal complement becomes ${a phi_1+b phi_2}$
    – reuns
    Nov 20 at 1:54












  • @reuns surely there is a condition on $a,b$? With $f(x) = begin{cases}x, & x < frac{1}{2}\ 1-x, & x geq frac{1}{2}end{cases}$ and $g(x) = a + bx$ I compute $langle f, grangle_{H^1([0,1])} = frac{1}{4}a + frac{1}{8}b$ which is not always zero
    – Neal
    Nov 20 at 2:04












  • That's why I gave two different inner products, only for the first one the orthogonal complement has an obvious basis, for the other inner product, I'd say we should substract to $1$ and $x$ their projection on the trigonometric orthogonal basis of $H^1_0$
    – reuns
    Nov 20 at 2:13










  • @reuns Ah, I see, thank you. Your comments make good sense. (I would note that I specified the inner product in the question.)
    – Neal
    Nov 20 at 13:33


















For $H^1([0,1])$, say $langle f,g rangle_{H^1([0,1])} = f(0)overline{g(0)}+f(1)overline{g(1)}+langle f',g' rangle_{L^2([0,1])}$. $H^1_0([0,1]) = { f in H^1([0,1]),f(0) = f(1)=0}={ f in H^1([0,1]),f(0) = f(1)=langle f',1 rangle_{L^2([0,1])}=0}$ so the orthogonal complement is ${a+bx}$. When changing the inner product (for say $(f,g ) = langle f,g rangle_{L^2([0,1])}+langle f',g' rangle_{L^2([0,1])}$) the orthogonal complement becomes ${a phi_1+b phi_2}$
– reuns
Nov 20 at 1:54






For $H^1([0,1])$, say $langle f,g rangle_{H^1([0,1])} = f(0)overline{g(0)}+f(1)overline{g(1)}+langle f',g' rangle_{L^2([0,1])}$. $H^1_0([0,1]) = { f in H^1([0,1]),f(0) = f(1)=0}={ f in H^1([0,1]),f(0) = f(1)=langle f',1 rangle_{L^2([0,1])}=0}$ so the orthogonal complement is ${a+bx}$. When changing the inner product (for say $(f,g ) = langle f,g rangle_{L^2([0,1])}+langle f',g' rangle_{L^2([0,1])}$) the orthogonal complement becomes ${a phi_1+b phi_2}$
– reuns
Nov 20 at 1:54














@reuns surely there is a condition on $a,b$? With $f(x) = begin{cases}x, & x < frac{1}{2}\ 1-x, & x geq frac{1}{2}end{cases}$ and $g(x) = a + bx$ I compute $langle f, grangle_{H^1([0,1])} = frac{1}{4}a + frac{1}{8}b$ which is not always zero
– Neal
Nov 20 at 2:04






@reuns surely there is a condition on $a,b$? With $f(x) = begin{cases}x, & x < frac{1}{2}\ 1-x, & x geq frac{1}{2}end{cases}$ and $g(x) = a + bx$ I compute $langle f, grangle_{H^1([0,1])} = frac{1}{4}a + frac{1}{8}b$ which is not always zero
– Neal
Nov 20 at 2:04














That's why I gave two different inner products, only for the first one the orthogonal complement has an obvious basis, for the other inner product, I'd say we should substract to $1$ and $x$ their projection on the trigonometric orthogonal basis of $H^1_0$
– reuns
Nov 20 at 2:13




That's why I gave two different inner products, only for the first one the orthogonal complement has an obvious basis, for the other inner product, I'd say we should substract to $1$ and $x$ their projection on the trigonometric orthogonal basis of $H^1_0$
– reuns
Nov 20 at 2:13












@reuns Ah, I see, thank you. Your comments make good sense. (I would note that I specified the inner product in the question.)
– Neal
Nov 20 at 13:33






@reuns Ah, I see, thank you. Your comments make good sense. (I would note that I specified the inner product in the question.)
– Neal
Nov 20 at 13:33












1 Answer
1






active

oldest

votes


















3














As reuns has stated in the comments, the answer depends on the inner product you choose in $H^1(Omega)$.
Let us fix the most common choice
$$
(u,v)_{H^1(Omega)} = int_Omega nabla u cdot nabla v + u , v , mathrm{d}x.$$



In this case,
the orthogonal complement of $H_0^1(Omega)$ consists precisely
of the (weak) solutions $u in H^1(Omega)$ of
$$ -Delta u + u = 0$$
(without B.C.). Indeed, the weak formulation of this PDE is
$$(u,v)_{H^1(Omega)} = 0 quadforall v in H_0^1(Omega).$$



For different inner products, you get different PDEs.






share|cite|improve this answer





















  • Ah hah, this makes perfect sense. Does this PDE have a name? (Aside from "unit-eigenfunction equation with no boundary condition")
    – Neal
    Nov 24 at 1:11








  • 1




    I don't know if there is a special name of this PDE.
    – gerw
    Nov 24 at 6:03











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005808%2fwhat-is-the-orthogonal-complement-of-h1-0-in-h1%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









3














As reuns has stated in the comments, the answer depends on the inner product you choose in $H^1(Omega)$.
Let us fix the most common choice
$$
(u,v)_{H^1(Omega)} = int_Omega nabla u cdot nabla v + u , v , mathrm{d}x.$$



In this case,
the orthogonal complement of $H_0^1(Omega)$ consists precisely
of the (weak) solutions $u in H^1(Omega)$ of
$$ -Delta u + u = 0$$
(without B.C.). Indeed, the weak formulation of this PDE is
$$(u,v)_{H^1(Omega)} = 0 quadforall v in H_0^1(Omega).$$



For different inner products, you get different PDEs.






share|cite|improve this answer





















  • Ah hah, this makes perfect sense. Does this PDE have a name? (Aside from "unit-eigenfunction equation with no boundary condition")
    – Neal
    Nov 24 at 1:11








  • 1




    I don't know if there is a special name of this PDE.
    – gerw
    Nov 24 at 6:03
















3














As reuns has stated in the comments, the answer depends on the inner product you choose in $H^1(Omega)$.
Let us fix the most common choice
$$
(u,v)_{H^1(Omega)} = int_Omega nabla u cdot nabla v + u , v , mathrm{d}x.$$



In this case,
the orthogonal complement of $H_0^1(Omega)$ consists precisely
of the (weak) solutions $u in H^1(Omega)$ of
$$ -Delta u + u = 0$$
(without B.C.). Indeed, the weak formulation of this PDE is
$$(u,v)_{H^1(Omega)} = 0 quadforall v in H_0^1(Omega).$$



For different inner products, you get different PDEs.






share|cite|improve this answer





















  • Ah hah, this makes perfect sense. Does this PDE have a name? (Aside from "unit-eigenfunction equation with no boundary condition")
    – Neal
    Nov 24 at 1:11








  • 1




    I don't know if there is a special name of this PDE.
    – gerw
    Nov 24 at 6:03














3












3








3






As reuns has stated in the comments, the answer depends on the inner product you choose in $H^1(Omega)$.
Let us fix the most common choice
$$
(u,v)_{H^1(Omega)} = int_Omega nabla u cdot nabla v + u , v , mathrm{d}x.$$



In this case,
the orthogonal complement of $H_0^1(Omega)$ consists precisely
of the (weak) solutions $u in H^1(Omega)$ of
$$ -Delta u + u = 0$$
(without B.C.). Indeed, the weak formulation of this PDE is
$$(u,v)_{H^1(Omega)} = 0 quadforall v in H_0^1(Omega).$$



For different inner products, you get different PDEs.






share|cite|improve this answer












As reuns has stated in the comments, the answer depends on the inner product you choose in $H^1(Omega)$.
Let us fix the most common choice
$$
(u,v)_{H^1(Omega)} = int_Omega nabla u cdot nabla v + u , v , mathrm{d}x.$$



In this case,
the orthogonal complement of $H_0^1(Omega)$ consists precisely
of the (weak) solutions $u in H^1(Omega)$ of
$$ -Delta u + u = 0$$
(without B.C.). Indeed, the weak formulation of this PDE is
$$(u,v)_{H^1(Omega)} = 0 quadforall v in H_0^1(Omega).$$



For different inner products, you get different PDEs.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 20 at 7:20









gerw

19k11133




19k11133












  • Ah hah, this makes perfect sense. Does this PDE have a name? (Aside from "unit-eigenfunction equation with no boundary condition")
    – Neal
    Nov 24 at 1:11








  • 1




    I don't know if there is a special name of this PDE.
    – gerw
    Nov 24 at 6:03


















  • Ah hah, this makes perfect sense. Does this PDE have a name? (Aside from "unit-eigenfunction equation with no boundary condition")
    – Neal
    Nov 24 at 1:11








  • 1




    I don't know if there is a special name of this PDE.
    – gerw
    Nov 24 at 6:03
















Ah hah, this makes perfect sense. Does this PDE have a name? (Aside from "unit-eigenfunction equation with no boundary condition")
– Neal
Nov 24 at 1:11






Ah hah, this makes perfect sense. Does this PDE have a name? (Aside from "unit-eigenfunction equation with no boundary condition")
– Neal
Nov 24 at 1:11






1




1




I don't know if there is a special name of this PDE.
– gerw
Nov 24 at 6:03




I don't know if there is a special name of this PDE.
– gerw
Nov 24 at 6:03


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005808%2fwhat-is-the-orthogonal-complement-of-h1-0-in-h1%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

How to send String Array data to Server using php in android

Image
0 I have a String array which contains value string array = {6,37,38,837,7,...} and I have also used some string values. I want to send that all string values and string array to my database which is located at server side. When I am click on submit button the data will stored in table. On the PHP side I want to read this array as it is,other thing i will do at background means PHP side. See my below code then you will get it what I am says. Here is my code: public class Hotels extends Activity { // Declare Variables JSONArray jsonarray = null; private static final String TAG_ID = "id"; public static final String TAG_NAME = "name"; public static final String TAG_LOCATION = "location"; public static final String TAG_DESC = "description"; String f_date, l_date;...
Read more

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Image
up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
Read more

Is anime1.com a legal site for watching anime?

Image
up vote 4 down vote favorite 1 Other anime streaming sites are blocking my IP because I'm Iranian while anime1.com isn't. I wanted to know if it's legal to watch/stream anime from this site. Thanks. anime-production share | improve this question edited Nov 27 at 22:38 W. Are 1,599 2 23 asked Nov 27 at 20:36 mhnajafi 23 3 add a comment ...
Read more