Integrability of the Fourier transform in Sobolev space
$begingroup$
I believe that the statement below is a standard fact but I haven't figured out yet:
Suppose $fin L^{1}(mathbb{R}^{n})$ has integrable partial derivatives of order $n+1$ and $D^{alpha}fin L^{1}(mathbb{R}^{n})$ for all multiindex with $|alpha|le n+1$ (here $D^{alpha}$ is the mixed partial derivative). Prove that the Fourier transform $hat{f}$ is in $L^{1}(mathbb{R}^{n})$.
Does anyone know the proof or a reference for this statement?
sobolev-spaces fourier-transform harmonic-analysis
edited Dec 11 '18 at 17:16
Bernard
121k740116
asked Dec 11 '18 at 16:24
vutuanhienvutuanhien
568
$endgroup$
add a comment |
$begingroup$
I believe that the statement below is a standard fact but I haven't figured out yet:
Suppose $fin L^{1}(mathbb{R}^{n})$ has integrable partial derivatives of order $n+1$ and $D^{alpha}fin L^{1}(mathbb{R}^{n})$ for all multiindex with $|alpha|le n+1$ (here $D^{alpha}$ is the mixed partial derivative). Prove that the Fourier transform $hat{f}$ is in $L^{1}(mathbb{R}^{n})$.
Does anyone know the proof or a reference for this statement?
sobolev-spaces fourier-transform harmonic-analysis
edited Dec 11 '18 at 17:16
Bernard
121k740116
asked Dec 11 '18 at 16:24
vutuanhienvutuanhien
568
$endgroup$
2
$begingroup$
So $hat{f}(1+sum_k |xi_k|^{n+1})$ is bounded, where $1/(1+sum_k |xi_k|^{n+1}) in L^1(R^n)$
$endgroup$
– reuns
Dec 11 '18 at 16:57
$begingroup$
@reuns: My attempt: By Cauchy - Schwarz inequality, $left(int |hat{f}(xi)|right)^2le int 1/(1+|xi|^{n+1})^{2}int (1+|xi|^{n+1})^{2}hat{f}(xi)^{2}$ and the last integral is equal to $sum_{alphale n+1} hat{D^{alpha}f}(xi)$. Am I right?
$endgroup$
– vutuanhien
Dec 11 '18 at 18:54
$begingroup$
Who said the last integral converges ? I'm saying $|hat{f}(1+sum_k |xi_k|^{n+1})|le C implies |hat{f} |_{L^1} le C |1/(1+sum_k |xi_k|^{n+1}) |_{L^1}$. $C$ is obtained from the fact $partial_{xi_k}^{n+1} f in L^1$ so its Fourier transform is bounded
$endgroup$
– reuns
Dec 11 '18 at 18:57
add a comment |
$begingroup$
I believe that the statement below is a standard fact but I haven't figured out yet:
Suppose $fin L^{1}(mathbb{R}^{n})$ has integrable partial derivatives of order $n+1$ and $D^{alpha}fin L^{1}(mathbb{R}^{n})$ for all multiindex with $|alpha|le n+1$ (here $D^{alpha}$ is the mixed partial derivative). Prove that the Fourier transform $hat{f}$ is in $L^{1}(mathbb{R}^{n})$.
Does anyone know the proof or a reference for this statement?
sobolev-spaces fourier-transform harmonic-analysis
edited Dec 11 '18 at 17:16
Bernard
121k740116
asked Dec 11 '18 at 16:24
vutuanhienvutuanhien
568
$endgroup$
I believe that the statement below is a standard fact but I haven't figured out yet:
Suppose $fin L^{1}(mathbb{R}^{n})$ has integrable partial derivatives of order $n+1$ and $D^{alpha}fin L^{1}(mathbb{R}^{n})$ for all multiindex with $|alpha|le n+1$ (here $D^{alpha}$ is the mixed partial derivative). Prove that the Fourier transform $hat{f}$ is in $L^{1}(mathbb{R}^{n})$.
Does anyone know the proof or a reference for this statement?
sobolev-spaces fourier-transform harmonic-analysis
sobolev-spaces fourier-transform harmonic-analysis
edited Dec 11 '18 at 17:16
Bernard
121k740116
asked Dec 11 '18 at 16:24
vutuanhienvutuanhien
568
edited Dec 11 '18 at 17:16
Bernard
121k740116
asked Dec 11 '18 at 16:24
vutuanhienvutuanhien
568
edited Dec 11 '18 at 17:16
Bernard
121k740116
edited Dec 11 '18 at 17:16
Bernard
121k740116
edited Dec 11 '18 at 17:16
Bernard
121k740116
121k740116
asked Dec 11 '18 at 16:24
vutuanhienvutuanhien
568
asked Dec 11 '18 at 16:24
vutuanhienvutuanhien
568
asked Dec 11 '18 at 16:24
vutuanhienvutuanhien
568
568
2
$begingroup$
So $hat{f}(1+sum_k |xi_k|^{n+1})$ is bounded, where $1/(1+sum_k |xi_k|^{n+1}) in L^1(R^n)$
$endgroup$
– reuns
Dec 11 '18 at 16:57
$begingroup$
@reuns: My attempt: By Cauchy - Schwarz inequality, $left(int |hat{f}(xi)|right)^2le int 1/(1+|xi|^{n+1})^{2}int (1+|xi|^{n+1})^{2}hat{f}(xi)^{2}$ and the last integral is equal to $sum_{alphale n+1} hat{D^{alpha}f}(xi)$. Am I right?
$endgroup$
– vutuanhien
Dec 11 '18 at 18:54
$begingroup$
Who said the last integral converges ? I'm saying $|hat{f}(1+sum_k |xi_k|^{n+1})|le C implies |hat{f} |_{L^1} le C |1/(1+sum_k |xi_k|^{n+1}) |_{L^1}$. $C$ is obtained from the fact $partial_{xi_k}^{n+1} f in L^1$ so its Fourier transform is bounded
$endgroup$
– reuns
Dec 11 '18 at 18:57
add a comment |
2
$begingroup$
So $hat{f}(1+sum_k |xi_k|^{n+1})$ is bounded, where $1/(1+sum_k |xi_k|^{n+1}) in L^1(R^n)$
$endgroup$
– reuns
Dec 11 '18 at 16:57
$begingroup$
@reuns: My attempt: By Cauchy - Schwarz inequality, $left(int |hat{f}(xi)|right)^2le int 1/(1+|xi|^{n+1})^{2}int (1+|xi|^{n+1})^{2}hat{f}(xi)^{2}$ and the last integral is equal to $sum_{alphale n+1} hat{D^{alpha}f}(xi)$. Am I right?
$endgroup$
– vutuanhien
Dec 11 '18 at 18:54
$begingroup$
Who said the last integral converges ? I'm saying $|hat{f}(1+sum_k |xi_k|^{n+1})|le C implies |hat{f} |_{L^1} le C |1/(1+sum_k |xi_k|^{n+1}) |_{L^1}$. $C$ is obtained from the fact $partial_{xi_k}^{n+1} f in L^1$ so its Fourier transform is bounded
$endgroup$
– reuns
Dec 11 '18 at 18:57
2
2
$begingroup$
So $hat{f}(1+sum_k |xi_k|^{n+1})$ is bounded, where $1/(1+sum_k |xi_k|^{n+1}) in L^1(R^n)$
$endgroup$
– reuns
Dec 11 '18 at 16:57
$begingroup$
So $hat{f}(1+sum_k |xi_k|^{n+1})$ is bounded, where $1/(1+sum_k |xi_k|^{n+1}) in L^1(R^n)$
$endgroup$
– reuns
Dec 11 '18 at 16:57
$begingroup$
@reuns: My attempt: By Cauchy - Schwarz inequality, $left(int |hat{f}(xi)|right)^2le int 1/(1+|xi|^{n+1})^{2}int (1+|xi|^{n+1})^{2}hat{f}(xi)^{2}$ and the last integral is equal to $sum_{alphale n+1} hat{D^{alpha}f}(xi)$. Am I right?
$endgroup$
– vutuanhien
Dec 11 '18 at 18:54
$begingroup$
@reuns: My attempt: By Cauchy - Schwarz inequality, $left(int |hat{f}(xi)|right)^2le int 1/(1+|xi|^{n+1})^{2}int (1+|xi|^{n+1})^{2}hat{f}(xi)^{2}$ and the last integral is equal to $sum_{alphale n+1} hat{D^{alpha}f}(xi)$. Am I right?
$endgroup$
– vutuanhien
Dec 11 '18 at 18:54
$begingroup$
Who said the last integral converges ? I'm saying $|hat{f}(1+sum_k |xi_k|^{n+1})|le C implies |hat{f} |_{L^1} le C |1/(1+sum_k |xi_k|^{n+1}) |_{L^1}$. $C$ is obtained from the fact $partial_{xi_k}^{n+1} f in L^1$ so its Fourier transform is bounded
$endgroup$
– reuns
Dec 11 '18 at 18:57
$begingroup$
Who said the last integral converges ? I'm saying $|hat{f}(1+sum_k |xi_k|^{n+1})|le C implies |hat{f} |_{L^1} le C |1/(1+sum_k |xi_k|^{n+1}) |_{L^1}$. $C$ is obtained from the fact $partial_{xi_k}^{n+1} f in L^1$ so its Fourier transform is bounded
$endgroup$
– reuns
Dec 11 '18 at 18:57
add a comment |
0
active
oldest
votes
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%2f3035460%2fintegrability-of-the-fourier-transform-in-sobolev-space%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3035460%2fintegrability-of-the-fourier-transform-in-sobolev-space%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
2
$begingroup$
So $hat{f}(1+sum_k |xi_k|^{n+1})$ is bounded, where $1/(1+sum_k |xi_k|^{n+1}) in L^1(R^n)$
$endgroup$
– reuns
Dec 11 '18 at 16:57
$begingroup$
@reuns: My attempt: By Cauchy - Schwarz inequality, $left(int |hat{f}(xi)|right)^2le int 1/(1+|xi|^{n+1})^{2}int (1+|xi|^{n+1})^{2}hat{f}(xi)^{2}$ and the last integral is equal to $sum_{alphale n+1} hat{D^{alpha}f}(xi)$. Am I right?
$endgroup$
– vutuanhien
Dec 11 '18 at 18:54
$begingroup$
Who said the last integral converges ? I'm saying $|hat{f}(1+sum_k |xi_k|^{n+1})|le C implies |hat{f} |_{L^1} le C |1/(1+sum_k |xi_k|^{n+1}) |_{L^1}$. $C$ is obtained from the fact $partial_{xi_k}^{n+1} f in L^1$ so its Fourier transform is bounded
$endgroup$
– reuns
Dec 11 '18 at 18:57