Given that $f in L^2(mathbb{T})$ and the sequence of Fourier coefficients $(hat{f_n})in l^1(mathbb{Z})$, must...
3
Note that $mathbb{T} = { z in mathbb{C} : |z| = 1 }$. This detail wouldn't fit in the title.
This is a previous exam question I am practicing with and I'm at a loss! Any advice on how to think about this problem would be much appreciated.
functional-analysis fourier-series lp-spaces
asked Jan 4 at 1:11
kkckkc
1308
New contributor
kkc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
|
show 5 more comments
3
Note that $mathbb{T} = { z in mathbb{C} : |z| = 1 }$. This detail wouldn't fit in the title.
This is a previous exam question I am practicing with and I'm at a loss! Any advice on how to think about this problem would be much appreciated.
functional-analysis fourier-series lp-spaces
asked Jan 4 at 1:11
kkckkc
1308
New contributor
kkc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What is $hat{f}_n$?
– Ben W
Jan 4 at 1:15
2
$f = sum_n langle f,e_n rangle e_n$ in $L^2(mathbb{T})$. If $sum_n |langle f,e_n rangle| < infty$ then $sum_n langle f,e_n rangle e_n$ converges uniformly thus is continuous and $f$ is equal to a continuous function almost everywhere and in $L^2(mathbb{T})$
– reuns
Jan 4 at 1:26
2
@stressedout in the context of complex analysis it is standard to denote the unit circle by $mathbb{T}$.
– Ben W
Jan 4 at 1:32
1
Sure $(e_n)$ is the orthonormal basis of complex exponentials, $langle .,. rangle$ is the inner product and $langle f,e_n rangle = hat{f}(n)$ (or $hat{f}(n)/sqrt{2pi}$ depending on your normalization)
– reuns
Jan 4 at 3:59
2
@stressedout I was re-reading my old Fourier Analysis textbook and I remember better now so as to clear up your confusion. The unit circle is the 1-dimensional torus, which justifies the notation $mathbb{T}$. Also, we use the torus because functions on the torus are analogous to $2pi$-periodic functions on $mathbb{R}$, and these are apparently of great interest in Fourier analysis.
– Ben W
2 days ago
|
show 5 more comments
3
3
3
0
Note that $mathbb{T} = { z in mathbb{C} : |z| = 1 }$. This detail wouldn't fit in the title.
This is a previous exam question I am practicing with and I'm at a loss! Any advice on how to think about this problem would be much appreciated.
functional-analysis fourier-series lp-spaces
asked Jan 4 at 1:11
kkckkc
1308
New contributor
kkc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Note that $mathbb{T} = { z in mathbb{C} : |z| = 1 }$. This detail wouldn't fit in the title.
This is a previous exam question I am practicing with and I'm at a loss! Any advice on how to think about this problem would be much appreciated.
functional-analysis fourier-series lp-spaces
functional-analysis fourier-series lp-spaces
asked Jan 4 at 1:11
kkckkc
1308
New contributor
kkc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 4 at 1:11
kkckkc
1308
New contributor
kkc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 4 at 1:20
kkc
asked Jan 4 at 1:11
kkckkc
1308
New contributor
kkc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 4 at 1:11
kkckkc
1308
asked Jan 4 at 1:11
kkckkc
1308
1308
New contributor
kkc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
kkc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
kkc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What is $hat{f}_n$?
– Ben W
Jan 4 at 1:15
2
$f = sum_n langle f,e_n rangle e_n$ in $L^2(mathbb{T})$. If $sum_n |langle f,e_n rangle| < infty$ then $sum_n langle f,e_n rangle e_n$ converges uniformly thus is continuous and $f$ is equal to a continuous function almost everywhere and in $L^2(mathbb{T})$
– reuns
Jan 4 at 1:26
2
@stressedout in the context of complex analysis it is standard to denote the unit circle by $mathbb{T}$.
– Ben W
Jan 4 at 1:32
1
Sure $(e_n)$ is the orthonormal basis of complex exponentials, $langle .,. rangle$ is the inner product and $langle f,e_n rangle = hat{f}(n)$ (or $hat{f}(n)/sqrt{2pi}$ depending on your normalization)
– reuns
Jan 4 at 3:59
2
@stressedout I was re-reading my old Fourier Analysis textbook and I remember better now so as to clear up your confusion. The unit circle is the 1-dimensional torus, which justifies the notation $mathbb{T}$. Also, we use the torus because functions on the torus are analogous to $2pi$-periodic functions on $mathbb{R}$, and these are apparently of great interest in Fourier analysis.
– Ben W
2 days ago
|
show 5 more comments
What is $hat{f}_n$?
– Ben W
Jan 4 at 1:15
2
$f = sum_n langle f,e_n rangle e_n$ in $L^2(mathbb{T})$. If $sum_n |langle f,e_n rangle| < infty$ then $sum_n langle f,e_n rangle e_n$ converges uniformly thus is continuous and $f$ is equal to a continuous function almost everywhere and in $L^2(mathbb{T})$
– reuns
Jan 4 at 1:26
2
@stressedout in the context of complex analysis it is standard to denote the unit circle by $mathbb{T}$.
– Ben W
Jan 4 at 1:32
1
Sure $(e_n)$ is the orthonormal basis of complex exponentials, $langle .,. rangle$ is the inner product and $langle f,e_n rangle = hat{f}(n)$ (or $hat{f}(n)/sqrt{2pi}$ depending on your normalization)
– reuns
Jan 4 at 3:59
2
@stressedout I was re-reading my old Fourier Analysis textbook and I remember better now so as to clear up your confusion. The unit circle is the 1-dimensional torus, which justifies the notation $mathbb{T}$. Also, we use the torus because functions on the torus are analogous to $2pi$-periodic functions on $mathbb{R}$, and these are apparently of great interest in Fourier analysis.
– Ben W
2 days ago
What is $hat{f}_n$?
– Ben W
Jan 4 at 1:15
What is $hat{f}_n$?
– Ben W
Jan 4 at 1:15
2
2
$f = sum_n langle f,e_n rangle e_n$ in $L^2(mathbb{T})$. If $sum_n |langle f,e_n rangle| < infty$ then $sum_n langle f,e_n rangle e_n$ converges uniformly thus is continuous and $f$ is equal to a continuous function almost everywhere and in $L^2(mathbb{T})$
– reuns
Jan 4 at 1:26
$f = sum_n langle f,e_n rangle e_n$ in $L^2(mathbb{T})$. If $sum_n |langle f,e_n rangle| < infty$ then $sum_n langle f,e_n rangle e_n$ converges uniformly thus is continuous and $f$ is equal to a continuous function almost everywhere and in $L^2(mathbb{T})$
– reuns
Jan 4 at 1:26
2
2
@stressedout in the context of complex analysis it is standard to denote the unit circle by $mathbb{T}$.
– Ben W
Jan 4 at 1:32
@stressedout in the context of complex analysis it is standard to denote the unit circle by $mathbb{T}$.
– Ben W
Jan 4 at 1:32
1
1
Sure $(e_n)$ is the orthonormal basis of complex exponentials, $langle .,. rangle$ is the inner product and $langle f,e_n rangle = hat{f}(n)$ (or $hat{f}(n)/sqrt{2pi}$ depending on your normalization)
– reuns
Jan 4 at 3:59
Sure $(e_n)$ is the orthonormal basis of complex exponentials, $langle .,. rangle$ is the inner product and $langle f,e_n rangle = hat{f}(n)$ (or $hat{f}(n)/sqrt{2pi}$ depending on your normalization)
– reuns
Jan 4 at 3:59
2
2
@stressedout I was re-reading my old Fourier Analysis textbook and I remember better now so as to clear up your confusion. The unit circle is the 1-dimensional torus, which justifies the notation $mathbb{T}$. Also, we use the torus because functions on the torus are analogous to $2pi$-periodic functions on $mathbb{R}$, and these are apparently of great interest in Fourier analysis.
– Ben W
2 days ago
@stressedout I was re-reading my old Fourier Analysis textbook and I remember better now so as to clear up your confusion. The unit circle is the 1-dimensional torus, which justifies the notation $mathbb{T}$. Also, we use the torus because functions on the torus are analogous to $2pi$-periodic functions on $mathbb{R}$, and these are apparently of great interest in Fourier analysis.
– Ben W
2 days ago
|
show 5 more comments
1 Answer
1
active
oldest
votes
2
If $fin L^2$, then $sum_{n=-N}^{N}hat{f}(n)e^{inx}$ converges to $f$ in $L^2$. Assuming that $hat{f} in ell^1$, then the series also converges absolutely and uniformly, which means that $f$ is equal a.e. to a continuous function.
answered Jan 4 at 4:12
DisintegratingByPartsDisintegratingByParts
58.7k42579
Why does the fact that $sum_{n=-N}^N hat{f}(n) e^{inx}$ converges to $f$ in $L^2$ and uniformly imply that $f$ is continuous a.e.?
– kkc
Jan 4 at 4:43
Actually, I take it back. It's because $sum_{n=-N}^N hat{f}(n) e^{inx}$ is continuous and the uniform limit of a continuous function is continuous.
– kkc
Jan 4 at 4:45
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
});
}
});
kkc is a new contributor. Be nice, and check out our Code of Conduct.
draft saved
draft discarded
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%2f3061215%2fgiven-that-f-in-l2-mathbbt-and-the-sequence-of-fourier-coefficients-h%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
2
If $fin L^2$, then $sum_{n=-N}^{N}hat{f}(n)e^{inx}$ converges to $f$ in $L^2$. Assuming that $hat{f} in ell^1$, then the series also converges absolutely and uniformly, which means that $f$ is equal a.e. to a continuous function.
answered Jan 4 at 4:12
DisintegratingByPartsDisintegratingByParts
58.7k42579
Why does the fact that $sum_{n=-N}^N hat{f}(n) e^{inx}$ converges to $f$ in $L^2$ and uniformly imply that $f$ is continuous a.e.?
– kkc
Jan 4 at 4:43
Actually, I take it back. It's because $sum_{n=-N}^N hat{f}(n) e^{inx}$ is continuous and the uniform limit of a continuous function is continuous.
– kkc
Jan 4 at 4:45
add a comment |
2
If $fin L^2$, then $sum_{n=-N}^{N}hat{f}(n)e^{inx}$ converges to $f$ in $L^2$. Assuming that $hat{f} in ell^1$, then the series also converges absolutely and uniformly, which means that $f$ is equal a.e. to a continuous function.
answered Jan 4 at 4:12
DisintegratingByPartsDisintegratingByParts
58.7k42579
Why does the fact that $sum_{n=-N}^N hat{f}(n) e^{inx}$ converges to $f$ in $L^2$ and uniformly imply that $f$ is continuous a.e.?
– kkc
Jan 4 at 4:43
Actually, I take it back. It's because $sum_{n=-N}^N hat{f}(n) e^{inx}$ is continuous and the uniform limit of a continuous function is continuous.
– kkc
Jan 4 at 4:45
add a comment |
2
2
2
If $fin L^2$, then $sum_{n=-N}^{N}hat{f}(n)e^{inx}$ converges to $f$ in $L^2$. Assuming that $hat{f} in ell^1$, then the series also converges absolutely and uniformly, which means that $f$ is equal a.e. to a continuous function.
answered Jan 4 at 4:12
DisintegratingByPartsDisintegratingByParts
58.7k42579
If $fin L^2$, then $sum_{n=-N}^{N}hat{f}(n)e^{inx}$ converges to $f$ in $L^2$. Assuming that $hat{f} in ell^1$, then the series also converges absolutely and uniformly, which means that $f$ is equal a.e. to a continuous function.
answered Jan 4 at 4:12
DisintegratingByPartsDisintegratingByParts
58.7k42579
answered Jan 4 at 4:12
DisintegratingByPartsDisintegratingByParts
58.7k42579
answered Jan 4 at 4:12
DisintegratingByPartsDisintegratingByParts
58.7k42579
answered Jan 4 at 4:12
DisintegratingByPartsDisintegratingByParts
58.7k42579
58.7k42579
Why does the fact that $sum_{n=-N}^N hat{f}(n) e^{inx}$ converges to $f$ in $L^2$ and uniformly imply that $f$ is continuous a.e.?
– kkc
Jan 4 at 4:43
Actually, I take it back. It's because $sum_{n=-N}^N hat{f}(n) e^{inx}$ is continuous and the uniform limit of a continuous function is continuous.
– kkc
Jan 4 at 4:45
add a comment |
Why does the fact that $sum_{n=-N}^N hat{f}(n) e^{inx}$ converges to $f$ in $L^2$ and uniformly imply that $f$ is continuous a.e.?
– kkc
Jan 4 at 4:43
Actually, I take it back. It's because $sum_{n=-N}^N hat{f}(n) e^{inx}$ is continuous and the uniform limit of a continuous function is continuous.
– kkc
Jan 4 at 4:45
Why does the fact that $sum_{n=-N}^N hat{f}(n) e^{inx}$ converges to $f$ in $L^2$ and uniformly imply that $f$ is continuous a.e.?
– kkc
Jan 4 at 4:43
Why does the fact that $sum_{n=-N}^N hat{f}(n) e^{inx}$ converges to $f$ in $L^2$ and uniformly imply that $f$ is continuous a.e.?
– kkc
Jan 4 at 4:43
Actually, I take it back. It's because $sum_{n=-N}^N hat{f}(n) e^{inx}$ is continuous and the uniform limit of a continuous function is continuous.
– kkc
Jan 4 at 4:45
Actually, I take it back. It's because $sum_{n=-N}^N hat{f}(n) e^{inx}$ is continuous and the uniform limit of a continuous function is continuous.
– kkc
Jan 4 at 4:45
add a comment |
kkc is a new contributor. Be nice, and check out our Code of Conduct.
draft saved
draft discarded
kkc is a new contributor. Be nice, and check out our Code of Conduct.
kkc is a new contributor. Be nice, and check out our Code of Conduct.
kkc is a new contributor. Be nice, and check out our Code of Conduct.
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
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%2f3061215%2fgiven-that-f-in-l2-mathbbt-and-the-sequence-of-fourier-coefficients-h%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
What is $hat{f}_n$?
– Ben W
Jan 4 at 1:15
2
$f = sum_n langle f,e_n rangle e_n$ in $L^2(mathbb{T})$. If $sum_n |langle f,e_n rangle| < infty$ then $sum_n langle f,e_n rangle e_n$ converges uniformly thus is continuous and $f$ is equal to a continuous function almost everywhere and in $L^2(mathbb{T})$
– reuns
Jan 4 at 1:26
2
@stressedout in the context of complex analysis it is standard to denote the unit circle by $mathbb{T}$.
– Ben W
Jan 4 at 1:32
1
Sure $(e_n)$ is the orthonormal basis of complex exponentials, $langle .,. rangle$ is the inner product and $langle f,e_n rangle = hat{f}(n)$ (or $hat{f}(n)/sqrt{2pi}$ depending on your normalization)
– reuns
Jan 4 at 3:59
2
@stressedout I was re-reading my old Fourier Analysis textbook and I remember better now so as to clear up your confusion. The unit circle is the 1-dimensional torus, which justifies the notation $mathbb{T}$. Also, we use the torus because functions on the torus are analogous to $2pi$-periodic functions on $mathbb{R}$, and these are apparently of great interest in Fourier analysis.
– Ben W
2 days ago