Mixing
Decay of the Fourier transform once again
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
I've seen many questions and answers about the relation between the smoothness of some function and the rate at which its Fourier transform decays. However, I haven't found the answer to the following question:
If $f in C^1(mathbbR) cap L^1(mathbbR)$, i. e. $f$ is continuously differentiable and integrable on $mathbbR$, is it true that its Fourier transform is $mathcalO(t^-1)$ at infinity?
Note that we don't require $f'$ to be integrable.
If $f'$ were integrable this could be shown by the relation $hatf(t) = it widehatf'(t)$ which in turn follows from integration by parts. In my situation however the fourier transform of $f'$ does not even exist (or does is it in some sense?).
I've thought about approximating the original function $f$ by functions with compact support or Schwartz-functions but the best I can get by this method is that $f(t) leq C cdot t^-1 + epsilon$ for every $epsilon > 0$ and for some constant $C > 0$ which depends on $epsilon$.
real-analysis fourier-analysis
asked Jul 20 at 19:42
Bruno Krams
355
add a comment |Â
up vote
2
down vote
favorite
I've seen many questions and answers about the relation between the smoothness of some function and the rate at which its Fourier transform decays. However, I haven't found the answer to the following question:
If $f in C^1(mathbbR) cap L^1(mathbbR)$, i. e. $f$ is continuously differentiable and integrable on $mathbbR$, is it true that its Fourier transform is $mathcalO(t^-1)$ at infinity?
Note that we don't require $f'$ to be integrable.
If $f'$ were integrable this could be shown by the relation $hatf(t) = it widehatf'(t)$ which in turn follows from integration by parts. In my situation however the fourier transform of $f'$ does not even exist (or does is it in some sense?).
I've thought about approximating the original function $f$ by functions with compact support or Schwartz-functions but the best I can get by this method is that $f(t) leq C cdot t^-1 + epsilon$ for every $epsilon > 0$ and for some constant $C > 0$ which depends on $epsilon$.
real-analysis fourier-analysis
asked Jul 20 at 19:42
Bruno Krams
355
Do you mean that the Fourier transform decays as fast as $t^-1$?
– Gonzalo Benavides
Jul 20 at 19:46
@GonzaloBenavides Exactly.
– Bruno Krams
Jul 22 at 13:30
@Shaun Thanks for your clues. I've edited my question accordingly
– Bruno Krams
Jul 22 at 13:30
1
@Shaun Sorry, I was editing while I added the comment
– Bruno Krams
Jul 22 at 13:43
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I've seen many questions and answers about the relation between the smoothness of some function and the rate at which its Fourier transform decays. However, I haven't found the answer to the following question:
If $f in C^1(mathbbR) cap L^1(mathbbR)$, i. e. $f$ is continuously differentiable and integrable on $mathbbR$, is it true that its Fourier transform is $mathcalO(t^-1)$ at infinity?
Note that we don't require $f'$ to be integrable.
If $f'$ were integrable this could be shown by the relation $hatf(t) = it widehatf'(t)$ which in turn follows from integration by parts. In my situation however the fourier transform of $f'$ does not even exist (or does is it in some sense?).
I've thought about approximating the original function $f$ by functions with compact support or Schwartz-functions but the best I can get by this method is that $f(t) leq C cdot t^-1 + epsilon$ for every $epsilon > 0$ and for some constant $C > 0$ which depends on $epsilon$.
real-analysis fourier-analysis
asked Jul 20 at 19:42
Bruno Krams
355
I've seen many questions and answers about the relation between the smoothness of some function and the rate at which its Fourier transform decays. However, I haven't found the answer to the following question:
If $f in C^1(mathbbR) cap L^1(mathbbR)$, i. e. $f$ is continuously differentiable and integrable on $mathbbR$, is it true that its Fourier transform is $mathcalO(t^-1)$ at infinity?
Note that we don't require $f'$ to be integrable.
If $f'$ were integrable this could be shown by the relation $hatf(t) = it widehatf'(t)$ which in turn follows from integration by parts. In my situation however the fourier transform of $f'$ does not even exist (or does is it in some sense?).
I've thought about approximating the original function $f$ by functions with compact support or Schwartz-functions but the best I can get by this method is that $f(t) leq C cdot t^-1 + epsilon$ for every $epsilon > 0$ and for some constant $C > 0$ which depends on $epsilon$.
real-analysis fourier-analysis
asked Jul 20 at 19:42
Bruno Krams
355
edited Jul 22 at 13:44
asked Jul 20 at 19:42
Bruno Krams
355
asked Jul 20 at 19:42
Bruno Krams
355
asked Jul 20 at 19:42
Bruno Krams
355
355
Do you mean that the Fourier transform decays as fast as $t^-1$?
– Gonzalo Benavides
Jul 20 at 19:46
@GonzaloBenavides Exactly.
– Bruno Krams
Jul 22 at 13:30
@Shaun Thanks for your clues. I've edited my question accordingly
– Bruno Krams
Jul 22 at 13:30
1
@Shaun Sorry, I was editing while I added the comment
– Bruno Krams
Jul 22 at 13:43
add a comment |Â
Do you mean that the Fourier transform decays as fast as $t^-1$?
– Gonzalo Benavides
Jul 20 at 19:46
@GonzaloBenavides Exactly.
– Bruno Krams
Jul 22 at 13:30
@Shaun Thanks for your clues. I've edited my question accordingly
– Bruno Krams
Jul 22 at 13:30
1
@Shaun Sorry, I was editing while I added the comment
– Bruno Krams
Jul 22 at 13:43
Do you mean that the Fourier transform decays as fast as $t^-1$?
– Gonzalo Benavides
Jul 20 at 19:46
Do you mean that the Fourier transform decays as fast as $t^-1$?
– Gonzalo Benavides
Jul 20 at 19:46
@GonzaloBenavides Exactly.
– Bruno Krams
Jul 22 at 13:30
@GonzaloBenavides Exactly.
– Bruno Krams
Jul 22 at 13:30
@Shaun Thanks for your clues. I've edited my question accordingly
– Bruno Krams
Jul 22 at 13:30
@Shaun Thanks for your clues. I've edited my question accordingly
– Bruno Krams
Jul 22 at 13:30
1
1
@Shaun Sorry, I was editing while I added the comment
– Bruno Krams
Jul 22 at 13:43
@Shaun Sorry, I was editing while I added the comment
– Bruno Krams
Jul 22 at 13:43
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2857968%2fdecay-of-the-fourier-transform-once-again%23new-answer', 'question_page');
);
Post as a guest
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
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
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
Do you mean that the Fourier transform decays as fast as $t^-1$?
– Gonzalo Benavides
Jul 20 at 19:46
@GonzaloBenavides Exactly.
– Bruno Krams
Jul 22 at 13:30
@Shaun Thanks for your clues. I've edited my question accordingly
– Bruno Krams
Jul 22 at 13:30
1
@Shaun Sorry, I was editing while I added the comment
– Bruno Krams
Jul 22 at 13:43