Fourier Transform of Exponentially Decaying Function Cannot Have Compact Support
$begingroup$
Let $f: mathbb{R} rightarrow mathbb{R}$ be a measurable function, with $|f(x)| le e^{-|x|}$ a.e.
Then how can we prove that its Fourier transform, $hat{f}$, cannot have compact support (unless $f = 0$ a.e.).
I have a hint which says to show that $hat{f} in C^{infty}$; I can do this using the differentiation rules for Fourier transforms, but am unsure of how to proceed from here. Can we use this to show that $hat{f}$ is analytic in some neighbourhood of $mathbb{R}$ (in which case the result follows easily)?
I have another hint which says to then consider a suitable Taylor expansion.
I am not too sure what to make of the second hint, but answers that involve the hints would be preferred.
complex-analysis fourier-analysis fourier-transform distribution-theory analyticity
asked Dec 29 '18 at 12:52
John DonJohn Don
361115
$endgroup$
|
show 2 more comments
$begingroup$
Let $f: mathbb{R} rightarrow mathbb{R}$ be a measurable function, with $|f(x)| le e^{-|x|}$ a.e.
Then how can we prove that its Fourier transform, $hat{f}$, cannot have compact support (unless $f = 0$ a.e.).
I have a hint which says to show that $hat{f} in C^{infty}$; I can do this using the differentiation rules for Fourier transforms, but am unsure of how to proceed from here. Can we use this to show that $hat{f}$ is analytic in some neighbourhood of $mathbb{R}$ (in which case the result follows easily)?
I have another hint which says to then consider a suitable Taylor expansion.
I am not too sure what to make of the second hint, but answers that involve the hints would be preferred.
complex-analysis fourier-analysis fourier-transform distribution-theory analyticity
asked Dec 29 '18 at 12:52
John DonJohn Don
361115
$endgroup$
$begingroup$
How does the result follow easily assuming $f$ is analytic in some nbhd?
$endgroup$
– mathworker21
Dec 29 '18 at 13:01
$begingroup$
@mathworker21 Essentially by the identity theorem. Alternatively, see this question (if $f$ is analytic in a neighbourhood of $mathbb{R}$, then in particular, it is $mathbb{R}$-analytic).
$endgroup$
– John Don
Dec 29 '18 at 13:11
$begingroup$
Going back to the definition of the Fourier transform, you can show directly that $hat{f}$ extends to an analytic function in ${Im(z) >-1}$ (or ${Im(z) < 1}$ depending on which convention you use). @JohnDon
$endgroup$
– Michh
Dec 29 '18 at 13:12
1
$begingroup$
This is what you are looking for with the roles of $f$ and $hat{f}$ interchanged.
$endgroup$
– Michh
Dec 29 '18 at 13:31
$begingroup$
@JohnDon do you mean $hat{f}$ is analytic in some nbhd?
$endgroup$
– mathworker21
Dec 29 '18 at 13:34
|
show 2 more comments
$begingroup$
Let $f: mathbb{R} rightarrow mathbb{R}$ be a measurable function, with $|f(x)| le e^{-|x|}$ a.e.
Then how can we prove that its Fourier transform, $hat{f}$, cannot have compact support (unless $f = 0$ a.e.).
I have a hint which says to show that $hat{f} in C^{infty}$; I can do this using the differentiation rules for Fourier transforms, but am unsure of how to proceed from here. Can we use this to show that $hat{f}$ is analytic in some neighbourhood of $mathbb{R}$ (in which case the result follows easily)?
I have another hint which says to then consider a suitable Taylor expansion.
I am not too sure what to make of the second hint, but answers that involve the hints would be preferred.
complex-analysis fourier-analysis fourier-transform distribution-theory analyticity
asked Dec 29 '18 at 12:52
John DonJohn Don
361115
$endgroup$
Let $f: mathbb{R} rightarrow mathbb{R}$ be a measurable function, with $|f(x)| le e^{-|x|}$ a.e.
Then how can we prove that its Fourier transform, $hat{f}$, cannot have compact support (unless $f = 0$ a.e.).
I have a hint which says to show that $hat{f} in C^{infty}$; I can do this using the differentiation rules for Fourier transforms, but am unsure of how to proceed from here. Can we use this to show that $hat{f}$ is analytic in some neighbourhood of $mathbb{R}$ (in which case the result follows easily)?
I have another hint which says to then consider a suitable Taylor expansion.
I am not too sure what to make of the second hint, but answers that involve the hints would be preferred.
complex-analysis fourier-analysis fourier-transform distribution-theory analyticity
complex-analysis fourier-analysis fourier-transform distribution-theory analyticity
asked Dec 29 '18 at 12:52
John DonJohn Don
361115
asked Dec 29 '18 at 12:52
John DonJohn Don
361115
edited Dec 29 '18 at 13:36
John Don
asked Dec 29 '18 at 12:52
John DonJohn Don
361115
asked Dec 29 '18 at 12:52
John DonJohn Don
361115
asked Dec 29 '18 at 12:52
John DonJohn Don
361115
361115
$begingroup$
How does the result follow easily assuming $f$ is analytic in some nbhd?
$endgroup$
– mathworker21
Dec 29 '18 at 13:01
$begingroup$
@mathworker21 Essentially by the identity theorem. Alternatively, see this question (if $f$ is analytic in a neighbourhood of $mathbb{R}$, then in particular, it is $mathbb{R}$-analytic).
$endgroup$
– John Don
Dec 29 '18 at 13:11
$begingroup$
Going back to the definition of the Fourier transform, you can show directly that $hat{f}$ extends to an analytic function in ${Im(z) >-1}$ (or ${Im(z) < 1}$ depending on which convention you use). @JohnDon
$endgroup$
– Michh
Dec 29 '18 at 13:12
1
$begingroup$
This is what you are looking for with the roles of $f$ and $hat{f}$ interchanged.
$endgroup$
– Michh
Dec 29 '18 at 13:31
$begingroup$
@JohnDon do you mean $hat{f}$ is analytic in some nbhd?
$endgroup$
– mathworker21
Dec 29 '18 at 13:34
|
show 2 more comments
$begingroup$
How does the result follow easily assuming $f$ is analytic in some nbhd?
$endgroup$
– mathworker21
Dec 29 '18 at 13:01
$begingroup$
@mathworker21 Essentially by the identity theorem. Alternatively, see this question (if $f$ is analytic in a neighbourhood of $mathbb{R}$, then in particular, it is $mathbb{R}$-analytic).
$endgroup$
– John Don
Dec 29 '18 at 13:11
$begingroup$
Going back to the definition of the Fourier transform, you can show directly that $hat{f}$ extends to an analytic function in ${Im(z) >-1}$ (or ${Im(z) < 1}$ depending on which convention you use). @JohnDon
$endgroup$
– Michh
Dec 29 '18 at 13:12
1
$begingroup$
This is what you are looking for with the roles of $f$ and $hat{f}$ interchanged.
$endgroup$
– Michh
Dec 29 '18 at 13:31
$begingroup$
@JohnDon do you mean $hat{f}$ is analytic in some nbhd?
$endgroup$
– mathworker21
Dec 29 '18 at 13:34
$begingroup$
How does the result follow easily assuming $f$ is analytic in some nbhd?
$endgroup$
– mathworker21
Dec 29 '18 at 13:01
$begingroup$
How does the result follow easily assuming $f$ is analytic in some nbhd?
$endgroup$
– mathworker21
Dec 29 '18 at 13:01
$begingroup$
@mathworker21 Essentially by the identity theorem. Alternatively, see this question (if $f$ is analytic in a neighbourhood of $mathbb{R}$, then in particular, it is $mathbb{R}$-analytic).
$endgroup$
– John Don
Dec 29 '18 at 13:11
$begingroup$
@mathworker21 Essentially by the identity theorem. Alternatively, see this question (if $f$ is analytic in a neighbourhood of $mathbb{R}$, then in particular, it is $mathbb{R}$-analytic).
$endgroup$
– John Don
Dec 29 '18 at 13:11
$begingroup$
Going back to the definition of the Fourier transform, you can show directly that $hat{f}$ extends to an analytic function in ${Im(z) >-1}$ (or ${Im(z) < 1}$ depending on which convention you use). @JohnDon
$endgroup$
– Michh
Dec 29 '18 at 13:12
$begingroup$
Going back to the definition of the Fourier transform, you can show directly that $hat{f}$ extends to an analytic function in ${Im(z) >-1}$ (or ${Im(z) < 1}$ depending on which convention you use). @JohnDon
$endgroup$
– Michh
Dec 29 '18 at 13:12
1
1
$begingroup$
This is what you are looking for with the roles of $f$ and $hat{f}$ interchanged.
$endgroup$
– Michh
Dec 29 '18 at 13:31
$begingroup$
This is what you are looking for with the roles of $f$ and $hat{f}$ interchanged.
$endgroup$
– Michh
Dec 29 '18 at 13:31
$begingroup$
@JohnDon do you mean $hat{f}$ is analytic in some nbhd?
$endgroup$
– mathworker21
Dec 29 '18 at 13:34
$begingroup$
@JohnDon do you mean $hat{f}$ is analytic in some nbhd?
$endgroup$
– mathworker21
Dec 29 '18 at 13:34
|
show 2 more comments
1 Answer
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You can show that $hat{f}(s)$ is continuous in the strip $|Im s| < 1$. Dominated convergence will do the job. Then you can use Morera's theorem to show that $hat{f}$ is holomorphic in this strip by showing that $int_{Delta}hat{f}(s)ds=0$ for every triangle in the same strip; all this requires is interchanging orders of integration.
answered Jan 3 at 19:44
DisintegratingByPartsDisintegratingByParts
59.8k42681
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1 Answer
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1 Answer
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$begingroup$
You can show that $hat{f}(s)$ is continuous in the strip $|Im s| < 1$. Dominated convergence will do the job. Then you can use Morera's theorem to show that $hat{f}$ is holomorphic in this strip by showing that $int_{Delta}hat{f}(s)ds=0$ for every triangle in the same strip; all this requires is interchanging orders of integration.
answered Jan 3 at 19:44
DisintegratingByPartsDisintegratingByParts
59.8k42681
$endgroup$
add a comment |
$begingroup$
You can show that $hat{f}(s)$ is continuous in the strip $|Im s| < 1$. Dominated convergence will do the job. Then you can use Morera's theorem to show that $hat{f}$ is holomorphic in this strip by showing that $int_{Delta}hat{f}(s)ds=0$ for every triangle in the same strip; all this requires is interchanging orders of integration.
answered Jan 3 at 19:44
DisintegratingByPartsDisintegratingByParts
59.8k42681
$endgroup$
add a comment |
$begingroup$
You can show that $hat{f}(s)$ is continuous in the strip $|Im s| < 1$. Dominated convergence will do the job. Then you can use Morera's theorem to show that $hat{f}$ is holomorphic in this strip by showing that $int_{Delta}hat{f}(s)ds=0$ for every triangle in the same strip; all this requires is interchanging orders of integration.
answered Jan 3 at 19:44
DisintegratingByPartsDisintegratingByParts
59.8k42681
$endgroup$
You can show that $hat{f}(s)$ is continuous in the strip $|Im s| < 1$. Dominated convergence will do the job. Then you can use Morera's theorem to show that $hat{f}$ is holomorphic in this strip by showing that $int_{Delta}hat{f}(s)ds=0$ for every triangle in the same strip; all this requires is interchanging orders of integration.
answered Jan 3 at 19:44
DisintegratingByPartsDisintegratingByParts
59.8k42681
answered Jan 3 at 19:44
DisintegratingByPartsDisintegratingByParts
59.8k42681
answered Jan 3 at 19:44
DisintegratingByPartsDisintegratingByParts
59.8k42681
answered Jan 3 at 19:44
DisintegratingByPartsDisintegratingByParts
59.8k42681
59.8k42681
add a comment |
add a comment |
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$begingroup$
How does the result follow easily assuming $f$ is analytic in some nbhd?
$endgroup$
– mathworker21
Dec 29 '18 at 13:01
$begingroup$
@mathworker21 Essentially by the identity theorem. Alternatively, see this question (if $f$ is analytic in a neighbourhood of $mathbb{R}$, then in particular, it is $mathbb{R}$-analytic).
$endgroup$
– John Don
Dec 29 '18 at 13:11
$begingroup$
Going back to the definition of the Fourier transform, you can show directly that $hat{f}$ extends to an analytic function in ${Im(z) >-1}$ (or ${Im(z) < 1}$ depending on which convention you use). @JohnDon
$endgroup$
– Michh
Dec 29 '18 at 13:12
1
$begingroup$
This is what you are looking for with the roles of $f$ and $hat{f}$ interchanged.
$endgroup$
– Michh
Dec 29 '18 at 13:31
$begingroup$
@JohnDon do you mean $hat{f}$ is analytic in some nbhd?
$endgroup$
– mathworker21
Dec 29 '18 at 13:34