Fourier transform of $C^1$ class function
$begingroup$
I am struggle to prove the following theorem.
If $f$ $in$ $C^1$ and $f$ , $f'$ $in$ $L^1$ , then $mathcal{F}$$f$ $in$$L^1$.
($mathcal{F}$$f$ is Fourier transform of $f$ )
I doubt that whether this theorem is right.
Please tell me proof or counterexample.
(following statement is supplement)
I already know that $f$ is bounded , $mathcal{F}$$f$ $in$ $L^2$ and $f$ $in$ $L^2$ under the hypothesis.
And, since $f$, $f'$ $in$ $L^1$ then $mathcal{F}$$f$ , $mathcal{F}$$f'$ is bounded.
real-analysis fourier-analysis lebesgue-integral
edited Nov 29 '18 at 16:50
Davide Giraudo
125k16150261
asked Nov 29 '18 at 10:19
NikolaiNikolai
65
$endgroup$
add a comment |
$begingroup$
I am struggle to prove the following theorem.
If $f$ $in$ $C^1$ and $f$ , $f'$ $in$ $L^1$ , then $mathcal{F}$$f$ $in$$L^1$.
($mathcal{F}$$f$ is Fourier transform of $f$ )
I doubt that whether this theorem is right.
Please tell me proof or counterexample.
(following statement is supplement)
I already know that $f$ is bounded , $mathcal{F}$$f$ $in$ $L^2$ and $f$ $in$ $L^2$ under the hypothesis.
And, since $f$, $f'$ $in$ $L^1$ then $mathcal{F}$$f$ , $mathcal{F}$$f'$ is bounded.
real-analysis fourier-analysis lebesgue-integral
edited Nov 29 '18 at 16:50
Davide Giraudo
125k16150261
asked Nov 29 '18 at 10:19
NikolaiNikolai
65
$endgroup$
$begingroup$
$f' in L^2$ so $mathcal{F} [f'] = iomega hat{f} in L^2$ whence $langle 1/(1+|omega|),(1+|omega|) |hat{f}| rangle le ...$
$endgroup$
– reuns
Nov 29 '18 at 10:27
$begingroup$
@reuns Why f' is belong to L2 ?
$endgroup$
– Nikolai
Nov 29 '18 at 11:28
$begingroup$
@reuns If $f'$ is bounded, your inequality is correct. But, only $f'$ in $L^1$, can't lead $f'$ is bounded.
$endgroup$
– Nikolai
Nov 29 '18 at 12:02
$begingroup$
Sorry I missed the ambiguity : you should read $C^1$ as the Banach space with sup norm of $f$ and $f'$. If you read it as just "continuously differentiable" it is very possible the claim isn't true
$endgroup$
– reuns
Nov 29 '18 at 12:06
$begingroup$
Otherwise we can look at something like $f' ast |x|^{-1/2+epsilon}$ to obtain that $(1+| omega|^{1/2+epsilon}) hat{f} in L^2$
$endgroup$
– reuns
Nov 29 '18 at 12:15
add a comment |
$begingroup$
I am struggle to prove the following theorem.
If $f$ $in$ $C^1$ and $f$ , $f'$ $in$ $L^1$ , then $mathcal{F}$$f$ $in$$L^1$.
($mathcal{F}$$f$ is Fourier transform of $f$ )
I doubt that whether this theorem is right.
Please tell me proof or counterexample.
(following statement is supplement)
I already know that $f$ is bounded , $mathcal{F}$$f$ $in$ $L^2$ and $f$ $in$ $L^2$ under the hypothesis.
And, since $f$, $f'$ $in$ $L^1$ then $mathcal{F}$$f$ , $mathcal{F}$$f'$ is bounded.
real-analysis fourier-analysis lebesgue-integral
edited Nov 29 '18 at 16:50
Davide Giraudo
125k16150261
asked Nov 29 '18 at 10:19
NikolaiNikolai
65
$endgroup$
I am struggle to prove the following theorem.
If $f$ $in$ $C^1$ and $f$ , $f'$ $in$ $L^1$ , then $mathcal{F}$$f$ $in$$L^1$.
($mathcal{F}$$f$ is Fourier transform of $f$ )
I doubt that whether this theorem is right.
Please tell me proof or counterexample.
(following statement is supplement)
I already know that $f$ is bounded , $mathcal{F}$$f$ $in$ $L^2$ and $f$ $in$ $L^2$ under the hypothesis.
And, since $f$, $f'$ $in$ $L^1$ then $mathcal{F}$$f$ , $mathcal{F}$$f'$ is bounded.
real-analysis fourier-analysis lebesgue-integral
real-analysis fourier-analysis lebesgue-integral
edited Nov 29 '18 at 16:50
Davide Giraudo
125k16150261
asked Nov 29 '18 at 10:19
NikolaiNikolai
65
edited Nov 29 '18 at 16:50
Davide Giraudo
125k16150261
asked Nov 29 '18 at 10:19
NikolaiNikolai
65
edited Nov 29 '18 at 16:50
Davide Giraudo
125k16150261
edited Nov 29 '18 at 16:50
Davide Giraudo
125k16150261
edited Nov 29 '18 at 16:50
Davide Giraudo
125k16150261
125k16150261
asked Nov 29 '18 at 10:19
NikolaiNikolai
65
asked Nov 29 '18 at 10:19
NikolaiNikolai
65
asked Nov 29 '18 at 10:19
NikolaiNikolai
65
65
$begingroup$
$f' in L^2$ so $mathcal{F} [f'] = iomega hat{f} in L^2$ whence $langle 1/(1+|omega|),(1+|omega|) |hat{f}| rangle le ...$
$endgroup$
– reuns
Nov 29 '18 at 10:27
$begingroup$
@reuns Why f' is belong to L2 ?
$endgroup$
– Nikolai
Nov 29 '18 at 11:28
$begingroup$
@reuns If $f'$ is bounded, your inequality is correct. But, only $f'$ in $L^1$, can't lead $f'$ is bounded.
$endgroup$
– Nikolai
Nov 29 '18 at 12:02
$begingroup$
Sorry I missed the ambiguity : you should read $C^1$ as the Banach space with sup norm of $f$ and $f'$. If you read it as just "continuously differentiable" it is very possible the claim isn't true
$endgroup$
– reuns
Nov 29 '18 at 12:06
$begingroup$
Otherwise we can look at something like $f' ast |x|^{-1/2+epsilon}$ to obtain that $(1+| omega|^{1/2+epsilon}) hat{f} in L^2$
$endgroup$
– reuns
Nov 29 '18 at 12:15
add a comment |
$begingroup$
$f' in L^2$ so $mathcal{F} [f'] = iomega hat{f} in L^2$ whence $langle 1/(1+|omega|),(1+|omega|) |hat{f}| rangle le ...$
$endgroup$
– reuns
Nov 29 '18 at 10:27
$begingroup$
@reuns Why f' is belong to L2 ?
$endgroup$
– Nikolai
Nov 29 '18 at 11:28
$begingroup$
@reuns If $f'$ is bounded, your inequality is correct. But, only $f'$ in $L^1$, can't lead $f'$ is bounded.
$endgroup$
– Nikolai
Nov 29 '18 at 12:02
$begingroup$
Sorry I missed the ambiguity : you should read $C^1$ as the Banach space with sup norm of $f$ and $f'$. If you read it as just "continuously differentiable" it is very possible the claim isn't true
$endgroup$
– reuns
Nov 29 '18 at 12:06
$begingroup$
Otherwise we can look at something like $f' ast |x|^{-1/2+epsilon}$ to obtain that $(1+| omega|^{1/2+epsilon}) hat{f} in L^2$
$endgroup$
– reuns
Nov 29 '18 at 12:15
$begingroup$
$f' in L^2$ so $mathcal{F} [f'] = iomega hat{f} in L^2$ whence $langle 1/(1+|omega|),(1+|omega|) |hat{f}| rangle le ...$
$endgroup$
– reuns
Nov 29 '18 at 10:27
$begingroup$
$f' in L^2$ so $mathcal{F} [f'] = iomega hat{f} in L^2$ whence $langle 1/(1+|omega|),(1+|omega|) |hat{f}| rangle le ...$
$endgroup$
– reuns
Nov 29 '18 at 10:27
$begingroup$
@reuns Why f' is belong to L2 ?
$endgroup$
– Nikolai
Nov 29 '18 at 11:28
$begingroup$
@reuns Why f' is belong to L2 ?
$endgroup$
– Nikolai
Nov 29 '18 at 11:28
$begingroup$
@reuns If $f'$ is bounded, your inequality is correct. But, only $f'$ in $L^1$, can't lead $f'$ is bounded.
$endgroup$
– Nikolai
Nov 29 '18 at 12:02
$begingroup$
@reuns If $f'$ is bounded, your inequality is correct. But, only $f'$ in $L^1$, can't lead $f'$ is bounded.
$endgroup$
– Nikolai
Nov 29 '18 at 12:02
$begingroup$
Sorry I missed the ambiguity : you should read $C^1$ as the Banach space with sup norm of $f$ and $f'$. If you read it as just "continuously differentiable" it is very possible the claim isn't true
$endgroup$
– reuns
Nov 29 '18 at 12:06
$begingroup$
Sorry I missed the ambiguity : you should read $C^1$ as the Banach space with sup norm of $f$ and $f'$. If you read it as just "continuously differentiable" it is very possible the claim isn't true
$endgroup$
– reuns
Nov 29 '18 at 12:06
$begingroup$
Otherwise we can look at something like $f' ast |x|^{-1/2+epsilon}$ to obtain that $(1+| omega|^{1/2+epsilon}) hat{f} in L^2$
$endgroup$
– reuns
Nov 29 '18 at 12:15
$begingroup$
Otherwise we can look at something like $f' ast |x|^{-1/2+epsilon}$ to obtain that $(1+| omega|^{1/2+epsilon}) hat{f} in L^2$
$endgroup$
– reuns
Nov 29 '18 at 12:15
add a comment |
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$begingroup$
$f' in L^2$ so $mathcal{F} [f'] = iomega hat{f} in L^2$ whence $langle 1/(1+|omega|),(1+|omega|) |hat{f}| rangle le ...$
$endgroup$
– reuns
Nov 29 '18 at 10:27
$begingroup$
@reuns Why f' is belong to L2 ?
$endgroup$
– Nikolai
Nov 29 '18 at 11:28
$begingroup$
@reuns If $f'$ is bounded, your inequality is correct. But, only $f'$ in $L^1$, can't lead $f'$ is bounded.
$endgroup$
– Nikolai
Nov 29 '18 at 12:02
$begingroup$
Sorry I missed the ambiguity : you should read $C^1$ as the Banach space with sup norm of $f$ and $f'$. If you read it as just "continuously differentiable" it is very possible the claim isn't true
$endgroup$
– reuns
Nov 29 '18 at 12:06
$begingroup$
Otherwise we can look at something like $f' ast |x|^{-1/2+epsilon}$ to obtain that $(1+| omega|^{1/2+epsilon}) hat{f} in L^2$
$endgroup$
– reuns
Nov 29 '18 at 12:15