Fourier Transform of Exponentially Decaying Function Cannot Have Compact Support












2












$begingroup$


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







share|cite|improve this question











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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
















2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














2












2








2


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 13:36







John Don

















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


















  • $begingroup$
    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










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.






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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.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 3 at 19:44









        DisintegratingByPartsDisintegratingByParts

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