Find the orthogonal complement of a subset of $L^2$












2












$begingroup$


I can't find the orthogonal of this set $C= left{ u in L^2(0,2): int_0^2 u(t)dt=1 right}$.




By applying the standard definition I have to find some conditions on $v(x)$ such that $int_0^2 u(x)v(x)dx=0$ for every $u in L^2$. I tried to write $v(x)=v(x)+1-1$ in order to use the fact that $u$ belongs to $C$, but I still can't solve the problem. For sure $v(x)=0 in C^{perp}$



Any hint or else would be really appreciated






functional-analysis hilbert-spaces lp-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    I would set an orthonormal basis, such that Legendre polynomials, write the functions in such basis and see what happens with the coefficients.
    $endgroup$
    – Dog_69
    Dec 27 '18 at 18:38












  • $begingroup$
    Sorry but I don't know anything about Legendre Polynomials yet
    $endgroup$
    – VoB
    Dec 27 '18 at 18:50
















2












$begingroup$


I can't find the orthogonal of this set $C= left{ u in L^2(0,2): int_0^2 u(t)dt=1 right}$.




By applying the standard definition I have to find some conditions on $v(x)$ such that $int_0^2 u(x)v(x)dx=0$ for every $u in L^2$. I tried to write $v(x)=v(x)+1-1$ in order to use the fact that $u$ belongs to $C$, but I still can't solve the problem. For sure $v(x)=0 in C^{perp}$



Any hint or else would be really appreciated






functional-analysis hilbert-spaces lp-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    I would set an orthonormal basis, such that Legendre polynomials, write the functions in such basis and see what happens with the coefficients.
    $endgroup$
    – Dog_69
    Dec 27 '18 at 18:38












  • $begingroup$
    Sorry but I don't know anything about Legendre Polynomials yet
    $endgroup$
    – VoB
    Dec 27 '18 at 18:50














2












2








2





$begingroup$


I can't find the orthogonal of this set $C= left{ u in L^2(0,2): int_0^2 u(t)dt=1 right}$.




By applying the standard definition I have to find some conditions on $v(x)$ such that $int_0^2 u(x)v(x)dx=0$ for every $u in L^2$. I tried to write $v(x)=v(x)+1-1$ in order to use the fact that $u$ belongs to $C$, but I still can't solve the problem. For sure $v(x)=0 in C^{perp}$



Any hint or else would be really appreciated






functional-analysis hilbert-spaces lp-spaces







share|cite|improve this question











$endgroup$




I can't find the orthogonal of this set $C= left{ u in L^2(0,2): int_0^2 u(t)dt=1 right}$.




By applying the standard definition I have to find some conditions on $v(x)$ such that $int_0^2 u(x)v(x)dx=0$ for every $u in L^2$. I tried to write $v(x)=v(x)+1-1$ in order to use the fact that $u$ belongs to $C$, but I still can't solve the problem. For sure $v(x)=0 in C^{perp}$



Any hint or else would be really appreciated






functional-analysis hilbert-spaces lp-spaces




functional-analysis hilbert-spaces lp-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 14:34









Davide Giraudo

128k17156268




128k17156268










asked Dec 27 '18 at 18:16









VoBVoB

801513




801513












  • $begingroup$
    I would set an orthonormal basis, such that Legendre polynomials, write the functions in such basis and see what happens with the coefficients.
    $endgroup$
    – Dog_69
    Dec 27 '18 at 18:38












  • $begingroup$
    Sorry but I don't know anything about Legendre Polynomials yet
    $endgroup$
    – VoB
    Dec 27 '18 at 18:50


















  • $begingroup$
    I would set an orthonormal basis, such that Legendre polynomials, write the functions in such basis and see what happens with the coefficients.
    $endgroup$
    – Dog_69
    Dec 27 '18 at 18:38












  • $begingroup$
    Sorry but I don't know anything about Legendre Polynomials yet
    $endgroup$
    – VoB
    Dec 27 '18 at 18:50
















$begingroup$
I would set an orthonormal basis, such that Legendre polynomials, write the functions in such basis and see what happens with the coefficients.
$endgroup$
– Dog_69
Dec 27 '18 at 18:38






$begingroup$
I would set an orthonormal basis, such that Legendre polynomials, write the functions in such basis and see what happens with the coefficients.
$endgroup$
– Dog_69
Dec 27 '18 at 18:38














$begingroup$
Sorry but I don't know anything about Legendre Polynomials yet
$endgroup$
– VoB
Dec 27 '18 at 18:50




$begingroup$
Sorry but I don't know anything about Legendre Polynomials yet
$endgroup$
– VoB
Dec 27 '18 at 18:50










1 Answer
1






active

oldest

votes


















5












$begingroup$

The orthogonal is trivial.



Indeed, assume $v in C^perp$.



If $int_0^2 v(t),dt ne 0$, then $$int_0^2 frac{v(x)}{int_0^2 v(t),dt},dx = frac{int_0^2 v(x),dx}{int_0^2 v(t),dt} = 1$$
so we have $frac{v}{int_0^2 v(t),dt} in C$. Therefore $v perp frac{v}{int_0^2 v(t),dt}$ which implies $v perp v$ and hence $v = 0$.



If $int_0^2 v(t),dt = 0$, then for any $u in C$ we have $$int_0^2 (u(t) + v(t)),dt = int_0^2 u(t),dt + int_0^2 v(t),dt = 1$$
so $u+v in C$. Hence $v perp u$ and $v perp u+v$ imply $v perp v$. It follows $v = 0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It's clear until you state "Hence $v perp u$" in the last line. From previous computations you showed that $u+v in C$, under the assumption $v in C^{perp}$.
    $endgroup$
    – VoB
    Dec 27 '18 at 19:11












  • $begingroup$
    Oh okay, you also had $u in C$... it's clear. Thanks for the answer
    $endgroup$
    – VoB
    Dec 27 '18 at 19:12












Your Answer








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%2f3054229%2ffind-the-orthogonal-complement-of-a-subset-of-l2%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









5












$begingroup$

The orthogonal is trivial.



Indeed, assume $v in C^perp$.



If $int_0^2 v(t),dt ne 0$, then $$int_0^2 frac{v(x)}{int_0^2 v(t),dt},dx = frac{int_0^2 v(x),dx}{int_0^2 v(t),dt} = 1$$
so we have $frac{v}{int_0^2 v(t),dt} in C$. Therefore $v perp frac{v}{int_0^2 v(t),dt}$ which implies $v perp v$ and hence $v = 0$.



If $int_0^2 v(t),dt = 0$, then for any $u in C$ we have $$int_0^2 (u(t) + v(t)),dt = int_0^2 u(t),dt + int_0^2 v(t),dt = 1$$
so $u+v in C$. Hence $v perp u$ and $v perp u+v$ imply $v perp v$. It follows $v = 0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It's clear until you state "Hence $v perp u$" in the last line. From previous computations you showed that $u+v in C$, under the assumption $v in C^{perp}$.
    $endgroup$
    – VoB
    Dec 27 '18 at 19:11












  • $begingroup$
    Oh okay, you also had $u in C$... it's clear. Thanks for the answer
    $endgroup$
    – VoB
    Dec 27 '18 at 19:12
















5












$begingroup$

The orthogonal is trivial.



Indeed, assume $v in C^perp$.



If $int_0^2 v(t),dt ne 0$, then $$int_0^2 frac{v(x)}{int_0^2 v(t),dt},dx = frac{int_0^2 v(x),dx}{int_0^2 v(t),dt} = 1$$
so we have $frac{v}{int_0^2 v(t),dt} in C$. Therefore $v perp frac{v}{int_0^2 v(t),dt}$ which implies $v perp v$ and hence $v = 0$.



If $int_0^2 v(t),dt = 0$, then for any $u in C$ we have $$int_0^2 (u(t) + v(t)),dt = int_0^2 u(t),dt + int_0^2 v(t),dt = 1$$
so $u+v in C$. Hence $v perp u$ and $v perp u+v$ imply $v perp v$. It follows $v = 0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It's clear until you state "Hence $v perp u$" in the last line. From previous computations you showed that $u+v in C$, under the assumption $v in C^{perp}$.
    $endgroup$
    – VoB
    Dec 27 '18 at 19:11












  • $begingroup$
    Oh okay, you also had $u in C$... it's clear. Thanks for the answer
    $endgroup$
    – VoB
    Dec 27 '18 at 19:12














5












5








5





$begingroup$

The orthogonal is trivial.



Indeed, assume $v in C^perp$.



If $int_0^2 v(t),dt ne 0$, then $$int_0^2 frac{v(x)}{int_0^2 v(t),dt},dx = frac{int_0^2 v(x),dx}{int_0^2 v(t),dt} = 1$$
so we have $frac{v}{int_0^2 v(t),dt} in C$. Therefore $v perp frac{v}{int_0^2 v(t),dt}$ which implies $v perp v$ and hence $v = 0$.



If $int_0^2 v(t),dt = 0$, then for any $u in C$ we have $$int_0^2 (u(t) + v(t)),dt = int_0^2 u(t),dt + int_0^2 v(t),dt = 1$$
so $u+v in C$. Hence $v perp u$ and $v perp u+v$ imply $v perp v$. It follows $v = 0$.






share|cite|improve this answer









$endgroup$



The orthogonal is trivial.



Indeed, assume $v in C^perp$.



If $int_0^2 v(t),dt ne 0$, then $$int_0^2 frac{v(x)}{int_0^2 v(t),dt},dx = frac{int_0^2 v(x),dx}{int_0^2 v(t),dt} = 1$$
so we have $frac{v}{int_0^2 v(t),dt} in C$. Therefore $v perp frac{v}{int_0^2 v(t),dt}$ which implies $v perp v$ and hence $v = 0$.



If $int_0^2 v(t),dt = 0$, then for any $u in C$ we have $$int_0^2 (u(t) + v(t)),dt = int_0^2 u(t),dt + int_0^2 v(t),dt = 1$$
so $u+v in C$. Hence $v perp u$ and $v perp u+v$ imply $v perp v$. It follows $v = 0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 27 '18 at 18:44









mechanodroidmechanodroid

28.9k62648




28.9k62648












  • $begingroup$
    It's clear until you state "Hence $v perp u$" in the last line. From previous computations you showed that $u+v in C$, under the assumption $v in C^{perp}$.
    $endgroup$
    – VoB
    Dec 27 '18 at 19:11












  • $begingroup$
    Oh okay, you also had $u in C$... it's clear. Thanks for the answer
    $endgroup$
    – VoB
    Dec 27 '18 at 19:12


















  • $begingroup$
    It's clear until you state "Hence $v perp u$" in the last line. From previous computations you showed that $u+v in C$, under the assumption $v in C^{perp}$.
    $endgroup$
    – VoB
    Dec 27 '18 at 19:11












  • $begingroup$
    Oh okay, you also had $u in C$... it's clear. Thanks for the answer
    $endgroup$
    – VoB
    Dec 27 '18 at 19:12
















$begingroup$
It's clear until you state "Hence $v perp u$" in the last line. From previous computations you showed that $u+v in C$, under the assumption $v in C^{perp}$.
$endgroup$
– VoB
Dec 27 '18 at 19:11






$begingroup$
It's clear until you state "Hence $v perp u$" in the last line. From previous computations you showed that $u+v in C$, under the assumption $v in C^{perp}$.
$endgroup$
– VoB
Dec 27 '18 at 19:11














$begingroup$
Oh okay, you also had $u in C$... it's clear. Thanks for the answer
$endgroup$
– VoB
Dec 27 '18 at 19:12




$begingroup$
Oh okay, you also had $u in C$... it's clear. Thanks for the answer
$endgroup$
– VoB
Dec 27 '18 at 19:12


















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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054229%2ffind-the-orthogonal-complement-of-a-subset-of-l2%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 change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
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