Fourier Transform - Duality formula. What are the necessary conditions?
$begingroup$
I've come across two contradicting statements which I'd be glad if you could help me resolve:
Theorem: if $fleft( x right)$ is continuous and absolutely integrable ($intlimits_{ - infty }^infty {left| {f(x)} right|dx} < infty $) and suppose $widehat fleft( omega right) = intlimits_{ - infty }^infty {fleft( x right){e^{ - iomega x}}dx} $ is absolutely integrable ($intlimits_{ - infty }^{ - infty } {{{left| {widehat fleft( omega right)} right|}}domega } < infty $) then $$Fleft{ {Fleft{ {fleft( x right)} right}} right} = 2pi fleft( { - x} right)$$
Just to clarify notation: $Fleft{ {fleft( x right)} right} = widehat fleft( omega right)$.
So this theorem assumes $f(x)$ is continuous on the real line.
But I really think we can demand less, that $f(x)$ be only piecewise continuous and get the continuity of $f(x)$ as a result of this theorem.
My reasoning:
if $f(x)$ is piecewise continuous and absolutely integrable then we know its fourier transform is continuous.
By the same thinking, since $Fleft{ {fleft( x right)} right}$ is continuous and absolutely integrable then its fourier transform, $2pi fleft( { - x} right)$, is also continuous.
Am I correct?
fourier-analysis fourier-transform
asked Dec 6 '18 at 6:35
zokomokozokomoko
174214
$endgroup$
|
show 2 more comments
$begingroup$
I've come across two contradicting statements which I'd be glad if you could help me resolve:
Theorem: if $fleft( x right)$ is continuous and absolutely integrable ($intlimits_{ - infty }^infty {left| {f(x)} right|dx} < infty $) and suppose $widehat fleft( omega right) = intlimits_{ - infty }^infty {fleft( x right){e^{ - iomega x}}dx} $ is absolutely integrable ($intlimits_{ - infty }^{ - infty } {{{left| {widehat fleft( omega right)} right|}}domega } < infty $) then $$Fleft{ {Fleft{ {fleft( x right)} right}} right} = 2pi fleft( { - x} right)$$
Just to clarify notation: $Fleft{ {fleft( x right)} right} = widehat fleft( omega right)$.
So this theorem assumes $f(x)$ is continuous on the real line.
But I really think we can demand less, that $f(x)$ be only piecewise continuous and get the continuity of $f(x)$ as a result of this theorem.
My reasoning:
if $f(x)$ is piecewise continuous and absolutely integrable then we know its fourier transform is continuous.
By the same thinking, since $Fleft{ {fleft( x right)} right}$ is continuous and absolutely integrable then its fourier transform, $2pi fleft( { - x} right)$, is also continuous.
Am I correct?
fourier-analysis fourier-transform
asked Dec 6 '18 at 6:35
zokomokozokomoko
174214
$endgroup$
$begingroup$
Yes. But the Fourier inversion theorem for $f,hat{f} in L^1$ is what you need to prove the rest.
$endgroup$
– reuns
Dec 7 '18 at 22:00
$begingroup$
You mean that I first need to prove that if $f,widehat f in {L^1}$ then $fleft( x right) = intlimits_{ - infty }^infty {widehat fleft( omega right){e^{iomega x}}domega } $? If so, should that equality hold pointwise or in the ${L^1}$ sense?
$endgroup$
– zokomoko
Dec 7 '18 at 22:16
$begingroup$
$lim_{n to infty}intlimits_{ - infty }^infty {widehat fleft( omega right)e^{-omega^2/n^2}e^{iomega x}domega }$ converges to $2pi f$ in quite all the normed space where $f$ belongs to. When $widehat{f} in L^1$ we can easily link it to $lim_{n to infty}intlimits_{ -n}^n {widehat fleft( omega right)e^{iomega x}domega }$ and obtain uniform convergence, $L^2$ convergence ...
$endgroup$
– reuns
Dec 8 '18 at 20:30
$begingroup$
What are the domains $mathbb{R},mathbb{C} ?$ and range $mathbb{R},mathbb{C} ?$ of your $f$ ?
$endgroup$
– Jean Marie
Dec 14 '18 at 18:34
$begingroup$
A rather vague suggestion : you can pass from a piecewise linear function to a continuous fonction by convolving it with a continuous fonction. Thus I wouldn't be astonished that an explanation stems out of a certain convolution (known to be Fourier-friendly). But which one ? By the limit of a certain kernel ?
$endgroup$
– Jean Marie
Dec 14 '18 at 23:17
|
show 2 more comments
$begingroup$
I've come across two contradicting statements which I'd be glad if you could help me resolve:
Theorem: if $fleft( x right)$ is continuous and absolutely integrable ($intlimits_{ - infty }^infty {left| {f(x)} right|dx} < infty $) and suppose $widehat fleft( omega right) = intlimits_{ - infty }^infty {fleft( x right){e^{ - iomega x}}dx} $ is absolutely integrable ($intlimits_{ - infty }^{ - infty } {{{left| {widehat fleft( omega right)} right|}}domega } < infty $) then $$Fleft{ {Fleft{ {fleft( x right)} right}} right} = 2pi fleft( { - x} right)$$
Just to clarify notation: $Fleft{ {fleft( x right)} right} = widehat fleft( omega right)$.
So this theorem assumes $f(x)$ is continuous on the real line.
But I really think we can demand less, that $f(x)$ be only piecewise continuous and get the continuity of $f(x)$ as a result of this theorem.
My reasoning:
if $f(x)$ is piecewise continuous and absolutely integrable then we know its fourier transform is continuous.
By the same thinking, since $Fleft{ {fleft( x right)} right}$ is continuous and absolutely integrable then its fourier transform, $2pi fleft( { - x} right)$, is also continuous.
Am I correct?
fourier-analysis fourier-transform
asked Dec 6 '18 at 6:35
zokomokozokomoko
174214
$endgroup$
I've come across two contradicting statements which I'd be glad if you could help me resolve:
Theorem: if $fleft( x right)$ is continuous and absolutely integrable ($intlimits_{ - infty }^infty {left| {f(x)} right|dx} < infty $) and suppose $widehat fleft( omega right) = intlimits_{ - infty }^infty {fleft( x right){e^{ - iomega x}}dx} $ is absolutely integrable ($intlimits_{ - infty }^{ - infty } {{{left| {widehat fleft( omega right)} right|}}domega } < infty $) then $$Fleft{ {Fleft{ {fleft( x right)} right}} right} = 2pi fleft( { - x} right)$$
Just to clarify notation: $Fleft{ {fleft( x right)} right} = widehat fleft( omega right)$.
So this theorem assumes $f(x)$ is continuous on the real line.
But I really think we can demand less, that $f(x)$ be only piecewise continuous and get the continuity of $f(x)$ as a result of this theorem.
My reasoning:
if $f(x)$ is piecewise continuous and absolutely integrable then we know its fourier transform is continuous.
By the same thinking, since $Fleft{ {fleft( x right)} right}$ is continuous and absolutely integrable then its fourier transform, $2pi fleft( { - x} right)$, is also continuous.
Am I correct?
fourier-analysis fourier-transform
fourier-analysis fourier-transform
asked Dec 6 '18 at 6:35
zokomokozokomoko
174214
asked Dec 6 '18 at 6:35
zokomokozokomoko
174214
edited Dec 7 '18 at 16:46
zokomoko
asked Dec 6 '18 at 6:35
zokomokozokomoko
174214
asked Dec 6 '18 at 6:35
zokomokozokomoko
174214
asked Dec 6 '18 at 6:35
zokomokozokomoko
174214
174214
$begingroup$
Yes. But the Fourier inversion theorem for $f,hat{f} in L^1$ is what you need to prove the rest.
$endgroup$
– reuns
Dec 7 '18 at 22:00
$begingroup$
You mean that I first need to prove that if $f,widehat f in {L^1}$ then $fleft( x right) = intlimits_{ - infty }^infty {widehat fleft( omega right){e^{iomega x}}domega } $? If so, should that equality hold pointwise or in the ${L^1}$ sense?
$endgroup$
– zokomoko
Dec 7 '18 at 22:16
$begingroup$
$lim_{n to infty}intlimits_{ - infty }^infty {widehat fleft( omega right)e^{-omega^2/n^2}e^{iomega x}domega }$ converges to $2pi f$ in quite all the normed space where $f$ belongs to. When $widehat{f} in L^1$ we can easily link it to $lim_{n to infty}intlimits_{ -n}^n {widehat fleft( omega right)e^{iomega x}domega }$ and obtain uniform convergence, $L^2$ convergence ...
$endgroup$
– reuns
Dec 8 '18 at 20:30
$begingroup$
What are the domains $mathbb{R},mathbb{C} ?$ and range $mathbb{R},mathbb{C} ?$ of your $f$ ?
$endgroup$
– Jean Marie
Dec 14 '18 at 18:34
$begingroup$
A rather vague suggestion : you can pass from a piecewise linear function to a continuous fonction by convolving it with a continuous fonction. Thus I wouldn't be astonished that an explanation stems out of a certain convolution (known to be Fourier-friendly). But which one ? By the limit of a certain kernel ?
$endgroup$
– Jean Marie
Dec 14 '18 at 23:17
|
show 2 more comments
$begingroup$
Yes. But the Fourier inversion theorem for $f,hat{f} in L^1$ is what you need to prove the rest.
$endgroup$
– reuns
Dec 7 '18 at 22:00
$begingroup$
You mean that I first need to prove that if $f,widehat f in {L^1}$ then $fleft( x right) = intlimits_{ - infty }^infty {widehat fleft( omega right){e^{iomega x}}domega } $? If so, should that equality hold pointwise or in the ${L^1}$ sense?
$endgroup$
– zokomoko
Dec 7 '18 at 22:16
$begingroup$
$lim_{n to infty}intlimits_{ - infty }^infty {widehat fleft( omega right)e^{-omega^2/n^2}e^{iomega x}domega }$ converges to $2pi f$ in quite all the normed space where $f$ belongs to. When $widehat{f} in L^1$ we can easily link it to $lim_{n to infty}intlimits_{ -n}^n {widehat fleft( omega right)e^{iomega x}domega }$ and obtain uniform convergence, $L^2$ convergence ...
$endgroup$
– reuns
Dec 8 '18 at 20:30
$begingroup$
What are the domains $mathbb{R},mathbb{C} ?$ and range $mathbb{R},mathbb{C} ?$ of your $f$ ?
$endgroup$
– Jean Marie
Dec 14 '18 at 18:34
$begingroup$
A rather vague suggestion : you can pass from a piecewise linear function to a continuous fonction by convolving it with a continuous fonction. Thus I wouldn't be astonished that an explanation stems out of a certain convolution (known to be Fourier-friendly). But which one ? By the limit of a certain kernel ?
$endgroup$
– Jean Marie
Dec 14 '18 at 23:17
$begingroup$
Yes. But the Fourier inversion theorem for $f,hat{f} in L^1$ is what you need to prove the rest.
$endgroup$
– reuns
Dec 7 '18 at 22:00
$begingroup$
Yes. But the Fourier inversion theorem for $f,hat{f} in L^1$ is what you need to prove the rest.
$endgroup$
– reuns
Dec 7 '18 at 22:00
$begingroup$
You mean that I first need to prove that if $f,widehat f in {L^1}$ then $fleft( x right) = intlimits_{ - infty }^infty {widehat fleft( omega right){e^{iomega x}}domega } $? If so, should that equality hold pointwise or in the ${L^1}$ sense?
$endgroup$
– zokomoko
Dec 7 '18 at 22:16
$begingroup$
You mean that I first need to prove that if $f,widehat f in {L^1}$ then $fleft( x right) = intlimits_{ - infty }^infty {widehat fleft( omega right){e^{iomega x}}domega } $? If so, should that equality hold pointwise or in the ${L^1}$ sense?
$endgroup$
– zokomoko
Dec 7 '18 at 22:16
$begingroup$
$lim_{n to infty}intlimits_{ - infty }^infty {widehat fleft( omega right)e^{-omega^2/n^2}e^{iomega x}domega }$ converges to $2pi f$ in quite all the normed space where $f$ belongs to. When $widehat{f} in L^1$ we can easily link it to $lim_{n to infty}intlimits_{ -n}^n {widehat fleft( omega right)e^{iomega x}domega }$ and obtain uniform convergence, $L^2$ convergence ...
$endgroup$
– reuns
Dec 8 '18 at 20:30
$begingroup$
$lim_{n to infty}intlimits_{ - infty }^infty {widehat fleft( omega right)e^{-omega^2/n^2}e^{iomega x}domega }$ converges to $2pi f$ in quite all the normed space where $f$ belongs to. When $widehat{f} in L^1$ we can easily link it to $lim_{n to infty}intlimits_{ -n}^n {widehat fleft( omega right)e^{iomega x}domega }$ and obtain uniform convergence, $L^2$ convergence ...
$endgroup$
– reuns
Dec 8 '18 at 20:30
$begingroup$
What are the domains $mathbb{R},mathbb{C} ?$ and range $mathbb{R},mathbb{C} ?$ of your $f$ ?
$endgroup$
– Jean Marie
Dec 14 '18 at 18:34
$begingroup$
What are the domains $mathbb{R},mathbb{C} ?$ and range $mathbb{R},mathbb{C} ?$ of your $f$ ?
$endgroup$
– Jean Marie
Dec 14 '18 at 18:34
$begingroup$
A rather vague suggestion : you can pass from a piecewise linear function to a continuous fonction by convolving it with a continuous fonction. Thus I wouldn't be astonished that an explanation stems out of a certain convolution (known to be Fourier-friendly). But which one ? By the limit of a certain kernel ?
$endgroup$
– Jean Marie
Dec 14 '18 at 23:17
$begingroup$
A rather vague suggestion : you can pass from a piecewise linear function to a continuous fonction by convolving it with a continuous fonction. Thus I wouldn't be astonished that an explanation stems out of a certain convolution (known to be Fourier-friendly). But which one ? By the limit of a certain kernel ?
$endgroup$
– Jean Marie
Dec 14 '18 at 23:17
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Suppose $f(x)= 0$ for $xne 0,$ $f(0)=1.$ Then $f$ is piecewise continuous on $mathbb R.$ Since $f=0$ a.e., we have $F[f]equiv 0,$ and hence $F[F[f]](x)equiv 0.$ Thus $F[F[f]](x)$ does not equal $2pi f(-x)$ everywhere.
If you view $f$ as an everywhere defined function, there's no way around this phenomenon. You can change $f$ on a set of measure $0$ but $F[f],$ and hence $F[F[f]],$ will not change.
But there is something you can say about Fourier inversion and piecewise continuous (PWC) functions. For simplicity suppose $f:mathbb Rto mathbb R$ is continuous on $(-infty,0)cup(0,infty)$ and $fin L^1.$ I'm allowing any sort of discontinuity at $0$ at this point.
Claim: If $F[f]in L^1,$ then $f$ has a removable singularity at $0.$
Proof: The well known inversion theorem for $L^1$ shows
$$f(x)= -frac{F[F[f])(-x)}{2pi}$$
for a.e. $x.$ Let $g$ be the function on the right; then $g$ is continuous everywhere. We then have $f=g$ a.e. everywhere on $(-infty,0).$ But two continuous functions that agree a.e. on $(-infty,0)$ actually agree everywhere on $(-infty,0).$ (Nice exercise) The same holds on $(0,infty).$ So we need only redefine $f(0)=g(0)$ and then $f= g$ everywhere. Thus $f$ has removable discontinuity at $0$ as claimed.
The claim extends to any finite number of discontinuities, with basically the same proof. It would also extend to a countable set of discontinuities if things are defined in the right way.
answered Dec 12 '18 at 19:56
zhw.zhw.
73.8k43175
$endgroup$
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%2f3028134%2ffourier-transform-duality-formula-what-are-the-necessary-conditions%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
$begingroup$
Suppose $f(x)= 0$ for $xne 0,$ $f(0)=1.$ Then $f$ is piecewise continuous on $mathbb R.$ Since $f=0$ a.e., we have $F[f]equiv 0,$ and hence $F[F[f]](x)equiv 0.$ Thus $F[F[f]](x)$ does not equal $2pi f(-x)$ everywhere.
If you view $f$ as an everywhere defined function, there's no way around this phenomenon. You can change $f$ on a set of measure $0$ but $F[f],$ and hence $F[F[f]],$ will not change.
But there is something you can say about Fourier inversion and piecewise continuous (PWC) functions. For simplicity suppose $f:mathbb Rto mathbb R$ is continuous on $(-infty,0)cup(0,infty)$ and $fin L^1.$ I'm allowing any sort of discontinuity at $0$ at this point.
Claim: If $F[f]in L^1,$ then $f$ has a removable singularity at $0.$
Proof: The well known inversion theorem for $L^1$ shows
$$f(x)= -frac{F[F[f])(-x)}{2pi}$$
for a.e. $x.$ Let $g$ be the function on the right; then $g$ is continuous everywhere. We then have $f=g$ a.e. everywhere on $(-infty,0).$ But two continuous functions that agree a.e. on $(-infty,0)$ actually agree everywhere on $(-infty,0).$ (Nice exercise) The same holds on $(0,infty).$ So we need only redefine $f(0)=g(0)$ and then $f= g$ everywhere. Thus $f$ has removable discontinuity at $0$ as claimed.
The claim extends to any finite number of discontinuities, with basically the same proof. It would also extend to a countable set of discontinuities if things are defined in the right way.
answered Dec 12 '18 at 19:56
zhw.zhw.
73.8k43175
$endgroup$
add a comment |
$begingroup$
Suppose $f(x)= 0$ for $xne 0,$ $f(0)=1.$ Then $f$ is piecewise continuous on $mathbb R.$ Since $f=0$ a.e., we have $F[f]equiv 0,$ and hence $F[F[f]](x)equiv 0.$ Thus $F[F[f]](x)$ does not equal $2pi f(-x)$ everywhere.
If you view $f$ as an everywhere defined function, there's no way around this phenomenon. You can change $f$ on a set of measure $0$ but $F[f],$ and hence $F[F[f]],$ will not change.
But there is something you can say about Fourier inversion and piecewise continuous (PWC) functions. For simplicity suppose $f:mathbb Rto mathbb R$ is continuous on $(-infty,0)cup(0,infty)$ and $fin L^1.$ I'm allowing any sort of discontinuity at $0$ at this point.
Claim: If $F[f]in L^1,$ then $f$ has a removable singularity at $0.$
Proof: The well known inversion theorem for $L^1$ shows
$$f(x)= -frac{F[F[f])(-x)}{2pi}$$
for a.e. $x.$ Let $g$ be the function on the right; then $g$ is continuous everywhere. We then have $f=g$ a.e. everywhere on $(-infty,0).$ But two continuous functions that agree a.e. on $(-infty,0)$ actually agree everywhere on $(-infty,0).$ (Nice exercise) The same holds on $(0,infty).$ So we need only redefine $f(0)=g(0)$ and then $f= g$ everywhere. Thus $f$ has removable discontinuity at $0$ as claimed.
The claim extends to any finite number of discontinuities, with basically the same proof. It would also extend to a countable set of discontinuities if things are defined in the right way.
answered Dec 12 '18 at 19:56
zhw.zhw.
73.8k43175
$endgroup$
add a comment |
$begingroup$
Suppose $f(x)= 0$ for $xne 0,$ $f(0)=1.$ Then $f$ is piecewise continuous on $mathbb R.$ Since $f=0$ a.e., we have $F[f]equiv 0,$ and hence $F[F[f]](x)equiv 0.$ Thus $F[F[f]](x)$ does not equal $2pi f(-x)$ everywhere.
If you view $f$ as an everywhere defined function, there's no way around this phenomenon. You can change $f$ on a set of measure $0$ but $F[f],$ and hence $F[F[f]],$ will not change.
But there is something you can say about Fourier inversion and piecewise continuous (PWC) functions. For simplicity suppose $f:mathbb Rto mathbb R$ is continuous on $(-infty,0)cup(0,infty)$ and $fin L^1.$ I'm allowing any sort of discontinuity at $0$ at this point.
Claim: If $F[f]in L^1,$ then $f$ has a removable singularity at $0.$
Proof: The well known inversion theorem for $L^1$ shows
$$f(x)= -frac{F[F[f])(-x)}{2pi}$$
for a.e. $x.$ Let $g$ be the function on the right; then $g$ is continuous everywhere. We then have $f=g$ a.e. everywhere on $(-infty,0).$ But two continuous functions that agree a.e. on $(-infty,0)$ actually agree everywhere on $(-infty,0).$ (Nice exercise) The same holds on $(0,infty).$ So we need only redefine $f(0)=g(0)$ and then $f= g$ everywhere. Thus $f$ has removable discontinuity at $0$ as claimed.
The claim extends to any finite number of discontinuities, with basically the same proof. It would also extend to a countable set of discontinuities if things are defined in the right way.
answered Dec 12 '18 at 19:56
zhw.zhw.
73.8k43175
$endgroup$
Suppose $f(x)= 0$ for $xne 0,$ $f(0)=1.$ Then $f$ is piecewise continuous on $mathbb R.$ Since $f=0$ a.e., we have $F[f]equiv 0,$ and hence $F[F[f]](x)equiv 0.$ Thus $F[F[f]](x)$ does not equal $2pi f(-x)$ everywhere.
If you view $f$ as an everywhere defined function, there's no way around this phenomenon. You can change $f$ on a set of measure $0$ but $F[f],$ and hence $F[F[f]],$ will not change.
But there is something you can say about Fourier inversion and piecewise continuous (PWC) functions. For simplicity suppose $f:mathbb Rto mathbb R$ is continuous on $(-infty,0)cup(0,infty)$ and $fin L^1.$ I'm allowing any sort of discontinuity at $0$ at this point.
Claim: If $F[f]in L^1,$ then $f$ has a removable singularity at $0.$
Proof: The well known inversion theorem for $L^1$ shows
$$f(x)= -frac{F[F[f])(-x)}{2pi}$$
for a.e. $x.$ Let $g$ be the function on the right; then $g$ is continuous everywhere. We then have $f=g$ a.e. everywhere on $(-infty,0).$ But two continuous functions that agree a.e. on $(-infty,0)$ actually agree everywhere on $(-infty,0).$ (Nice exercise) The same holds on $(0,infty).$ So we need only redefine $f(0)=g(0)$ and then $f= g$ everywhere. Thus $f$ has removable discontinuity at $0$ as claimed.
The claim extends to any finite number of discontinuities, with basically the same proof. It would also extend to a countable set of discontinuities if things are defined in the right way.
answered Dec 12 '18 at 19:56
zhw.zhw.
73.8k43175
answered Dec 12 '18 at 19:56
zhw.zhw.
73.8k43175
answered Dec 12 '18 at 19:56
zhw.zhw.
73.8k43175
answered Dec 12 '18 at 19:56
zhw.zhw.
73.8k43175
73.8k43175
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%2f3028134%2ffourier-transform-duality-formula-what-are-the-necessary-conditions%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
$begingroup$
Yes. But the Fourier inversion theorem for $f,hat{f} in L^1$ is what you need to prove the rest.
$endgroup$
– reuns
Dec 7 '18 at 22:00
$begingroup$
You mean that I first need to prove that if $f,widehat f in {L^1}$ then $fleft( x right) = intlimits_{ - infty }^infty {widehat fleft( omega right){e^{iomega x}}domega } $? If so, should that equality hold pointwise or in the ${L^1}$ sense?
$endgroup$
– zokomoko
Dec 7 '18 at 22:16
$begingroup$
$lim_{n to infty}intlimits_{ - infty }^infty {widehat fleft( omega right)e^{-omega^2/n^2}e^{iomega x}domega }$ converges to $2pi f$ in quite all the normed space where $f$ belongs to. When $widehat{f} in L^1$ we can easily link it to $lim_{n to infty}intlimits_{ -n}^n {widehat fleft( omega right)e^{iomega x}domega }$ and obtain uniform convergence, $L^2$ convergence ...
$endgroup$
– reuns
Dec 8 '18 at 20:30
$begingroup$
What are the domains $mathbb{R},mathbb{C} ?$ and range $mathbb{R},mathbb{C} ?$ of your $f$ ?
$endgroup$
– Jean Marie
Dec 14 '18 at 18:34
$begingroup$
A rather vague suggestion : you can pass from a piecewise linear function to a continuous fonction by convolving it with a continuous fonction. Thus I wouldn't be astonished that an explanation stems out of a certain convolution (known to be Fourier-friendly). But which one ? By the limit of a certain kernel ?
$endgroup$
– Jean Marie
Dec 14 '18 at 23:17