Mixing
Find the orthogonal complement on $L^2[0,1]$ of all polynomials.
$begingroup$
The problem:
Determine the orthogonal complement on $L^2[0,1]$ to all polynomials.
My approach and intuition thus far:
I know for sure intuitively that the orthogonal complement would just be the space containing the zero function. I have two approaches to the problem,
(1) I would use the Stone-Weierstrass Approximation theorem to show that the polynomials are dense in the space of functions on $C[0,1]$. Since $C[0,1]$ is dense in $L^2[0,1]$ the proof would follow by noting the property:
Given a Hilbert Space $V$, the orthogonal complement has the property that:
$V^perp=overline{V}^perp$.
So then I would have to find the orthogonal complement to all functions which is just going to be the zero function.
(2) The other approach I had was to consider the orthogonal basis of all polynomials on $L^2[0,1]$. Then, I would just find the orthogonal complement of the basis.
real-analysis hilbert-spaces banach-spaces lp-spaces
asked Jan 26 at 21:03
DarelDarel
1249
$endgroup$
|
show 3 more comments
$begingroup$
The problem:
Determine the orthogonal complement on $L^2[0,1]$ to all polynomials.
My approach and intuition thus far:
I know for sure intuitively that the orthogonal complement would just be the space containing the zero function. I have two approaches to the problem,
(1) I would use the Stone-Weierstrass Approximation theorem to show that the polynomials are dense in the space of functions on $C[0,1]$. Since $C[0,1]$ is dense in $L^2[0,1]$ the proof would follow by noting the property:
Given a Hilbert Space $V$, the orthogonal complement has the property that:
$V^perp=overline{V}^perp$.
So then I would have to find the orthogonal complement to all functions which is just going to be the zero function.
(2) The other approach I had was to consider the orthogonal basis of all polynomials on $L^2[0,1]$. Then, I would just find the orthogonal complement of the basis.
real-analysis hilbert-spaces banach-spaces lp-spaces
asked Jan 26 at 21:03
DarelDarel
1249
$endgroup$
$begingroup$
(1) Correct. You just have to be careful with the two different norms $|cdot|_infty$ and $|cdot|_2$. (2) What basis are you refering to?
$endgroup$
– amsmath
Jan 26 at 21:08
$begingroup$
For (2) I was referring to the orthogonal basis of all polynomials.
$endgroup$
– Darel
Jan 26 at 21:09
$begingroup$
Repetitions of what you already phrased don't bring us any further. The truth is that there doesn't exist an "orthonormal basis of all polynomials". I ask again: what do you mean?
$endgroup$
– amsmath
Jan 26 at 21:11
1
$begingroup$
The fact that you write "orthogonal basis of all polynomials" tells me that you have not quite understood what you are supposed to in the exercise. Hilbert spaces have orthogonal (better: orthonormal) bases. But the space of all polynomials is not a Hilbert space because it is not complete. It can be seen as a subspace of $L^2[0,1]$, but this subspace is not closed. You have to show that its closure (which is much bigger) is in fact $L^2[0,1]$. However, there are orthonormal bases of $L^2[0,1]$ consisting of polynomials. But showing that their span is dense is in general hard to do.
$endgroup$
– amsmath
Jan 26 at 21:34
1
$begingroup$
Yeah, and it's basically one line. You only have to show that the polynomials are dense in $L^2[0,1]$, meaning that you can approximate every $L^2$-function arbitrarily well by a polynomial in the $L^2$-norm.
$endgroup$
– amsmath
Jan 26 at 21:50
|
show 3 more comments
$begingroup$
The problem:
Determine the orthogonal complement on $L^2[0,1]$ to all polynomials.
My approach and intuition thus far:
I know for sure intuitively that the orthogonal complement would just be the space containing the zero function. I have two approaches to the problem,
(1) I would use the Stone-Weierstrass Approximation theorem to show that the polynomials are dense in the space of functions on $C[0,1]$. Since $C[0,1]$ is dense in $L^2[0,1]$ the proof would follow by noting the property:
Given a Hilbert Space $V$, the orthogonal complement has the property that:
$V^perp=overline{V}^perp$.
So then I would have to find the orthogonal complement to all functions which is just going to be the zero function.
(2) The other approach I had was to consider the orthogonal basis of all polynomials on $L^2[0,1]$. Then, I would just find the orthogonal complement of the basis.
real-analysis hilbert-spaces banach-spaces lp-spaces
asked Jan 26 at 21:03
DarelDarel
1249
$endgroup$
The problem:
Determine the orthogonal complement on $L^2[0,1]$ to all polynomials.
My approach and intuition thus far:
I know for sure intuitively that the orthogonal complement would just be the space containing the zero function. I have two approaches to the problem,
(1) I would use the Stone-Weierstrass Approximation theorem to show that the polynomials are dense in the space of functions on $C[0,1]$. Since $C[0,1]$ is dense in $L^2[0,1]$ the proof would follow by noting the property:
Given a Hilbert Space $V$, the orthogonal complement has the property that:
$V^perp=overline{V}^perp$.
So then I would have to find the orthogonal complement to all functions which is just going to be the zero function.
(2) The other approach I had was to consider the orthogonal basis of all polynomials on $L^2[0,1]$. Then, I would just find the orthogonal complement of the basis.
real-analysis hilbert-spaces banach-spaces lp-spaces
real-analysis hilbert-spaces banach-spaces lp-spaces
asked Jan 26 at 21:03
DarelDarel
1249
asked Jan 26 at 21:03
DarelDarel
1249
edited Jan 26 at 21:08
Darel
asked Jan 26 at 21:03
DarelDarel
1249
asked Jan 26 at 21:03
DarelDarel
1249
asked Jan 26 at 21:03
DarelDarel
1249
1249
$begingroup$
(1) Correct. You just have to be careful with the two different norms $|cdot|_infty$ and $|cdot|_2$. (2) What basis are you refering to?
$endgroup$
– amsmath
Jan 26 at 21:08
$begingroup$
For (2) I was referring to the orthogonal basis of all polynomials.
$endgroup$
– Darel
Jan 26 at 21:09
$begingroup$
Repetitions of what you already phrased don't bring us any further. The truth is that there doesn't exist an "orthonormal basis of all polynomials". I ask again: what do you mean?
$endgroup$
– amsmath
Jan 26 at 21:11
1
$begingroup$
The fact that you write "orthogonal basis of all polynomials" tells me that you have not quite understood what you are supposed to in the exercise. Hilbert spaces have orthogonal (better: orthonormal) bases. But the space of all polynomials is not a Hilbert space because it is not complete. It can be seen as a subspace of $L^2[0,1]$, but this subspace is not closed. You have to show that its closure (which is much bigger) is in fact $L^2[0,1]$. However, there are orthonormal bases of $L^2[0,1]$ consisting of polynomials. But showing that their span is dense is in general hard to do.
$endgroup$
– amsmath
Jan 26 at 21:34
1
$begingroup$
Yeah, and it's basically one line. You only have to show that the polynomials are dense in $L^2[0,1]$, meaning that you can approximate every $L^2$-function arbitrarily well by a polynomial in the $L^2$-norm.
$endgroup$
– amsmath
Jan 26 at 21:50
|
show 3 more comments
$begingroup$
(1) Correct. You just have to be careful with the two different norms $|cdot|_infty$ and $|cdot|_2$. (2) What basis are you refering to?
$endgroup$
– amsmath
Jan 26 at 21:08
$begingroup$
For (2) I was referring to the orthogonal basis of all polynomials.
$endgroup$
– Darel
Jan 26 at 21:09
$begingroup$
Repetitions of what you already phrased don't bring us any further. The truth is that there doesn't exist an "orthonormal basis of all polynomials". I ask again: what do you mean?
$endgroup$
– amsmath
Jan 26 at 21:11
1
$begingroup$
The fact that you write "orthogonal basis of all polynomials" tells me that you have not quite understood what you are supposed to in the exercise. Hilbert spaces have orthogonal (better: orthonormal) bases. But the space of all polynomials is not a Hilbert space because it is not complete. It can be seen as a subspace of $L^2[0,1]$, but this subspace is not closed. You have to show that its closure (which is much bigger) is in fact $L^2[0,1]$. However, there are orthonormal bases of $L^2[0,1]$ consisting of polynomials. But showing that their span is dense is in general hard to do.
$endgroup$
– amsmath
Jan 26 at 21:34
1
$begingroup$
Yeah, and it's basically one line. You only have to show that the polynomials are dense in $L^2[0,1]$, meaning that you can approximate every $L^2$-function arbitrarily well by a polynomial in the $L^2$-norm.
$endgroup$
– amsmath
Jan 26 at 21:50
$begingroup$
(1) Correct. You just have to be careful with the two different norms $|cdot|_infty$ and $|cdot|_2$. (2) What basis are you refering to?
$endgroup$
– amsmath
Jan 26 at 21:08
$begingroup$
(1) Correct. You just have to be careful with the two different norms $|cdot|_infty$ and $|cdot|_2$. (2) What basis are you refering to?
$endgroup$
– amsmath
Jan 26 at 21:08
$begingroup$
For (2) I was referring to the orthogonal basis of all polynomials.
$endgroup$
– Darel
Jan 26 at 21:09
$begingroup$
For (2) I was referring to the orthogonal basis of all polynomials.
$endgroup$
– Darel
Jan 26 at 21:09
$begingroup$
Repetitions of what you already phrased don't bring us any further. The truth is that there doesn't exist an "orthonormal basis of all polynomials". I ask again: what do you mean?
$endgroup$
– amsmath
Jan 26 at 21:11
$begingroup$
Repetitions of what you already phrased don't bring us any further. The truth is that there doesn't exist an "orthonormal basis of all polynomials". I ask again: what do you mean?
$endgroup$
– amsmath
Jan 26 at 21:11
1
1
$begingroup$
The fact that you write "orthogonal basis of all polynomials" tells me that you have not quite understood what you are supposed to in the exercise. Hilbert spaces have orthogonal (better: orthonormal) bases. But the space of all polynomials is not a Hilbert space because it is not complete. It can be seen as a subspace of $L^2[0,1]$, but this subspace is not closed. You have to show that its closure (which is much bigger) is in fact $L^2[0,1]$. However, there are orthonormal bases of $L^2[0,1]$ consisting of polynomials. But showing that their span is dense is in general hard to do.
$endgroup$
– amsmath
Jan 26 at 21:34
$begingroup$
The fact that you write "orthogonal basis of all polynomials" tells me that you have not quite understood what you are supposed to in the exercise. Hilbert spaces have orthogonal (better: orthonormal) bases. But the space of all polynomials is not a Hilbert space because it is not complete. It can be seen as a subspace of $L^2[0,1]$, but this subspace is not closed. You have to show that its closure (which is much bigger) is in fact $L^2[0,1]$. However, there are orthonormal bases of $L^2[0,1]$ consisting of polynomials. But showing that their span is dense is in general hard to do.
$endgroup$
– amsmath
Jan 26 at 21:34
1
1
$begingroup$
Yeah, and it's basically one line. You only have to show that the polynomials are dense in $L^2[0,1]$, meaning that you can approximate every $L^2$-function arbitrarily well by a polynomial in the $L^2$-norm.
$endgroup$
– amsmath
Jan 26 at 21:50
$begingroup$
Yeah, and it's basically one line. You only have to show that the polynomials are dense in $L^2[0,1]$, meaning that you can approximate every $L^2$-function arbitrarily well by a polynomial in the $L^2$-norm.
$endgroup$
– amsmath
Jan 26 at 21:50
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Going with route (1) would be the easiest way to go. If I understand what you mean by (2), i.e., finding a maximal orthonormal set of polynomials and showing that this set of polynomials has no orthogonal complement, doesn't really gain you anything; you still need to show that the orthogonal complement of a set is zero, and you don't gain anything by only considering only an orthonormal subset.
answered Jan 26 at 21:16
AweyganAweygan
14.6k21442
$endgroup$
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%2f3088754%2ffind-the-orthogonal-complement-on-l20-1-of-all-polynomials%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
$begingroup$
Going with route (1) would be the easiest way to go. If I understand what you mean by (2), i.e., finding a maximal orthonormal set of polynomials and showing that this set of polynomials has no orthogonal complement, doesn't really gain you anything; you still need to show that the orthogonal complement of a set is zero, and you don't gain anything by only considering only an orthonormal subset.
answered Jan 26 at 21:16
AweyganAweygan
14.6k21442
$endgroup$
add a comment |
$begingroup$
Going with route (1) would be the easiest way to go. If I understand what you mean by (2), i.e., finding a maximal orthonormal set of polynomials and showing that this set of polynomials has no orthogonal complement, doesn't really gain you anything; you still need to show that the orthogonal complement of a set is zero, and you don't gain anything by only considering only an orthonormal subset.
answered Jan 26 at 21:16
AweyganAweygan
14.6k21442
$endgroup$
add a comment |
$begingroup$
Going with route (1) would be the easiest way to go. If I understand what you mean by (2), i.e., finding a maximal orthonormal set of polynomials and showing that this set of polynomials has no orthogonal complement, doesn't really gain you anything; you still need to show that the orthogonal complement of a set is zero, and you don't gain anything by only considering only an orthonormal subset.
answered Jan 26 at 21:16
AweyganAweygan
14.6k21442
$endgroup$
Going with route (1) would be the easiest way to go. If I understand what you mean by (2), i.e., finding a maximal orthonormal set of polynomials and showing that this set of polynomials has no orthogonal complement, doesn't really gain you anything; you still need to show that the orthogonal complement of a set is zero, and you don't gain anything by only considering only an orthonormal subset.
answered Jan 26 at 21:16
AweyganAweygan
14.6k21442
answered Jan 26 at 21:16
AweyganAweygan
14.6k21442
answered Jan 26 at 21:16
AweyganAweygan
14.6k21442
answered Jan 26 at 21:16
AweyganAweygan
14.6k21442
14.6k21442
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.
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%2f3088754%2ffind-the-orthogonal-complement-on-l20-1-of-all-polynomials%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
$begingroup$
(1) Correct. You just have to be careful with the two different norms $|cdot|_infty$ and $|cdot|_2$. (2) What basis are you refering to?
$endgroup$
– amsmath
Jan 26 at 21:08
$begingroup$
For (2) I was referring to the orthogonal basis of all polynomials.
$endgroup$
– Darel
Jan 26 at 21:09
$begingroup$
Repetitions of what you already phrased don't bring us any further. The truth is that there doesn't exist an "orthonormal basis of all polynomials". I ask again: what do you mean?
$endgroup$
– amsmath
Jan 26 at 21:11
1
$begingroup$
The fact that you write "orthogonal basis of all polynomials" tells me that you have not quite understood what you are supposed to in the exercise. Hilbert spaces have orthogonal (better: orthonormal) bases. But the space of all polynomials is not a Hilbert space because it is not complete. It can be seen as a subspace of $L^2[0,1]$, but this subspace is not closed. You have to show that its closure (which is much bigger) is in fact $L^2[0,1]$. However, there are orthonormal bases of $L^2[0,1]$ consisting of polynomials. But showing that their span is dense is in general hard to do.
$endgroup$
– amsmath
Jan 26 at 21:34
1
$begingroup$
Yeah, and it's basically one line. You only have to show that the polynomials are dense in $L^2[0,1]$, meaning that you can approximate every $L^2$-function arbitrarily well by a polynomial in the $L^2$-norm.
$endgroup$
– amsmath
Jan 26 at 21:50