Decay of Fourier coefficients of infinitely differentiable functions
up vote
1
down vote
favorite
For a $C^n[-pi,pi]$ function $f$ we have that $|hat{f}(k)|in O(1/k^n)$. This implies that if $f$ is $C^infty[-pi,pi]$ then its $k-th$ Fourier coefficient decays faster than any $1/k^n$, $ngeq0.$ My questions:
$bullet$ There is a name for this order of convergence?(something like "Schwarz order" if there exists such a denomination)
$bullet$ Can't we say more than this? Y mean, there are a stronger result that states that these coefficients decay at an exponential rate (for example)?
fourier-analysis
edited Nov 22 at 16:57
David C. Ullrich
57.6k43891
add a comment |
up vote
1
down vote
favorite
For a $C^n[-pi,pi]$ function $f$ we have that $|hat{f}(k)|in O(1/k^n)$. This implies that if $f$ is $C^infty[-pi,pi]$ then its $k-th$ Fourier coefficient decays faster than any $1/k^n$, $ngeq0.$ My questions:
$bullet$ There is a name for this order of convergence?(something like "Schwarz order" if there exists such a denomination)
$bullet$ Can't we say more than this? Y mean, there are a stronger result that states that these coefficients decay at an exponential rate (for example)?
fourier-analysis
edited Nov 22 at 16:57
David C. Ullrich
57.6k43891
There is a bit more that can be said if $f(t)$ is analytic in a complex neighborhood of the real line, e.g. $f(t)=sqrt{1- msin(t)^2}$ with $|m|<1$. In that case, one indeed has exponentially-fast decay of the Fourier components in the desired sense. For discussion/proof of this result, see Trethefen and Weideman's 2014 SIAM excellent review article "The Exponentially Convergent Trapezoidal Rule" especially eq. (3-11) and Theorem 3.2.
– Semiclassical
May 26 '16 at 2:51
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
For a $C^n[-pi,pi]$ function $f$ we have that $|hat{f}(k)|in O(1/k^n)$. This implies that if $f$ is $C^infty[-pi,pi]$ then its $k-th$ Fourier coefficient decays faster than any $1/k^n$, $ngeq0.$ My questions:
$bullet$ There is a name for this order of convergence?(something like "Schwarz order" if there exists such a denomination)
$bullet$ Can't we say more than this? Y mean, there are a stronger result that states that these coefficients decay at an exponential rate (for example)?
fourier-analysis
edited Nov 22 at 16:57
David C. Ullrich
57.6k43891
For a $C^n[-pi,pi]$ function $f$ we have that $|hat{f}(k)|in O(1/k^n)$. This implies that if $f$ is $C^infty[-pi,pi]$ then its $k-th$ Fourier coefficient decays faster than any $1/k^n$, $ngeq0.$ My questions:
$bullet$ There is a name for this order of convergence?(something like "Schwarz order" if there exists such a denomination)
$bullet$ Can't we say more than this? Y mean, there are a stronger result that states that these coefficients decay at an exponential rate (for example)?
fourier-analysis
fourier-analysis
edited Nov 22 at 16:57
David C. Ullrich
57.6k43891
edited Nov 22 at 16:57
David C. Ullrich
57.6k43891
edited Nov 22 at 16:57
David C. Ullrich
57.6k43891
edited Nov 22 at 16:57
David C. Ullrich
57.6k43891
edited Nov 22 at 16:57
David C. Ullrich
57.6k43891
57.6k43891
asked May 25 '16 at 23:45
user335721
There is a bit more that can be said if $f(t)$ is analytic in a complex neighborhood of the real line, e.g. $f(t)=sqrt{1- msin(t)^2}$ with $|m|<1$. In that case, one indeed has exponentially-fast decay of the Fourier components in the desired sense. For discussion/proof of this result, see Trethefen and Weideman's 2014 SIAM excellent review article "The Exponentially Convergent Trapezoidal Rule" especially eq. (3-11) and Theorem 3.2.
– Semiclassical
May 26 '16 at 2:51
add a comment |
There is a bit more that can be said if $f(t)$ is analytic in a complex neighborhood of the real line, e.g. $f(t)=sqrt{1- msin(t)^2}$ with $|m|<1$. In that case, one indeed has exponentially-fast decay of the Fourier components in the desired sense. For discussion/proof of this result, see Trethefen and Weideman's 2014 SIAM excellent review article "The Exponentially Convergent Trapezoidal Rule" especially eq. (3-11) and Theorem 3.2.
– Semiclassical
May 26 '16 at 2:51
There is a bit more that can be said if $f(t)$ is analytic in a complex neighborhood of the real line, e.g. $f(t)=sqrt{1- msin(t)^2}$ with $|m|<1$. In that case, one indeed has exponentially-fast decay of the Fourier components in the desired sense. For discussion/proof of this result, see Trethefen and Weideman's 2014 SIAM excellent review article "The Exponentially Convergent Trapezoidal Rule" especially eq. (3-11) and Theorem 3.2.
– Semiclassical
May 26 '16 at 2:51
There is a bit more that can be said if $f(t)$ is analytic in a complex neighborhood of the real line, e.g. $f(t)=sqrt{1- msin(t)^2}$ with $|m|<1$. In that case, one indeed has exponentially-fast decay of the Fourier components in the desired sense. For discussion/proof of this result, see Trethefen and Weideman's 2014 SIAM excellent review article "The Exponentially Convergent Trapezoidal Rule" especially eq. (3-11) and Theorem 3.2.
– Semiclassical
May 26 '16 at 2:51
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Well first, what you say about $C^infty([-pi,pi])$ is not true. For example, $fin C^infty([-pi,pi])$ if $f(t)=t$, but the coefficients of that function are not $O(1/k^2)$.
Let's talk about $2pi$-periodic functions - that's what you meant, or what you should have meant. I'd just say the coefficients were rapidly decreasing, and specify what that meant if it mattered.
No, there's no more that can be said. It's easy to see that if $hat f(k)=O(1/k^{n+2})$ then $fin C^n$ (somewhat stronger results in this direction exist that are not quite as easy). Hence if $hat f(k)=O(1/k^n)$ for all $n$ then $fin C^infty$.
answered May 25 '16 at 23:56
David C. Ullrich
57.6k43891
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',
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%2f1800184%2fdecay-of-fourier-coefficients-of-infinitely-differentiable-functions%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
up vote
1
down vote
accepted
Well first, what you say about $C^infty([-pi,pi])$ is not true. For example, $fin C^infty([-pi,pi])$ if $f(t)=t$, but the coefficients of that function are not $O(1/k^2)$.
Let's talk about $2pi$-periodic functions - that's what you meant, or what you should have meant. I'd just say the coefficients were rapidly decreasing, and specify what that meant if it mattered.
No, there's no more that can be said. It's easy to see that if $hat f(k)=O(1/k^{n+2})$ then $fin C^n$ (somewhat stronger results in this direction exist that are not quite as easy). Hence if $hat f(k)=O(1/k^n)$ for all $n$ then $fin C^infty$.
answered May 25 '16 at 23:56
David C. Ullrich
57.6k43891
add a comment |
up vote
1
down vote
accepted
Well first, what you say about $C^infty([-pi,pi])$ is not true. For example, $fin C^infty([-pi,pi])$ if $f(t)=t$, but the coefficients of that function are not $O(1/k^2)$.
Let's talk about $2pi$-periodic functions - that's what you meant, or what you should have meant. I'd just say the coefficients were rapidly decreasing, and specify what that meant if it mattered.
No, there's no more that can be said. It's easy to see that if $hat f(k)=O(1/k^{n+2})$ then $fin C^n$ (somewhat stronger results in this direction exist that are not quite as easy). Hence if $hat f(k)=O(1/k^n)$ for all $n$ then $fin C^infty$.
answered May 25 '16 at 23:56
David C. Ullrich
57.6k43891
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Well first, what you say about $C^infty([-pi,pi])$ is not true. For example, $fin C^infty([-pi,pi])$ if $f(t)=t$, but the coefficients of that function are not $O(1/k^2)$.
Let's talk about $2pi$-periodic functions - that's what you meant, or what you should have meant. I'd just say the coefficients were rapidly decreasing, and specify what that meant if it mattered.
No, there's no more that can be said. It's easy to see that if $hat f(k)=O(1/k^{n+2})$ then $fin C^n$ (somewhat stronger results in this direction exist that are not quite as easy). Hence if $hat f(k)=O(1/k^n)$ for all $n$ then $fin C^infty$.
answered May 25 '16 at 23:56
David C. Ullrich
57.6k43891
Well first, what you say about $C^infty([-pi,pi])$ is not true. For example, $fin C^infty([-pi,pi])$ if $f(t)=t$, but the coefficients of that function are not $O(1/k^2)$.
Let's talk about $2pi$-periodic functions - that's what you meant, or what you should have meant. I'd just say the coefficients were rapidly decreasing, and specify what that meant if it mattered.
No, there's no more that can be said. It's easy to see that if $hat f(k)=O(1/k^{n+2})$ then $fin C^n$ (somewhat stronger results in this direction exist that are not quite as easy). Hence if $hat f(k)=O(1/k^n)$ for all $n$ then $fin C^infty$.
answered May 25 '16 at 23:56
David C. Ullrich
57.6k43891
answered May 25 '16 at 23:56
David C. Ullrich
57.6k43891
answered May 25 '16 at 23:56
David C. Ullrich
57.6k43891
answered May 25 '16 at 23:56
David C. Ullrich
57.6k43891
57.6k43891
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%2f1800184%2fdecay-of-fourier-coefficients-of-infinitely-differentiable-functions%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
There is a bit more that can be said if $f(t)$ is analytic in a complex neighborhood of the real line, e.g. $f(t)=sqrt{1- msin(t)^2}$ with $|m|<1$. In that case, one indeed has exponentially-fast decay of the Fourier components in the desired sense. For discussion/proof of this result, see Trethefen and Weideman's 2014 SIAM excellent review article "The Exponentially Convergent Trapezoidal Rule" especially eq. (3-11) and Theorem 3.2.
– Semiclassical
May 26 '16 at 2:51