Fourier Transform of Exponentially Decaying Function Cannot Have Compact Support












2














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
























  • How does the result follow easily assuming $f$ is analytic in some nbhd?
    – mathworker21
    Dec 29 '18 at 13:01












  • @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).
    – John Don
    Dec 29 '18 at 13:11












  • 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
    – Michh
    Dec 29 '18 at 13:12






  • 1




    This is what you are looking for with the roles of $f$ and $hat{f}$ interchanged.
    – Michh
    Dec 29 '18 at 13:31










  • @JohnDon do you mean $hat{f}$ is analytic in some nbhd?
    – mathworker21
    Dec 29 '18 at 13:34
















2














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
























  • How does the result follow easily assuming $f$ is analytic in some nbhd?
    – mathworker21
    Dec 29 '18 at 13:01












  • @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).
    – John Don
    Dec 29 '18 at 13:11












  • 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
    – Michh
    Dec 29 '18 at 13:12






  • 1




    This is what you are looking for with the roles of $f$ and $hat{f}$ interchanged.
    – Michh
    Dec 29 '18 at 13:31










  • @JohnDon do you mean $hat{f}$ is analytic in some nbhd?
    – mathworker21
    Dec 29 '18 at 13:34














2












2








2


1





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















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

























asked Dec 29 '18 at 12:52









John Don

338115




338115












  • How does the result follow easily assuming $f$ is analytic in some nbhd?
    – mathworker21
    Dec 29 '18 at 13:01












  • @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).
    – John Don
    Dec 29 '18 at 13:11












  • 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
    – Michh
    Dec 29 '18 at 13:12






  • 1




    This is what you are looking for with the roles of $f$ and $hat{f}$ interchanged.
    – Michh
    Dec 29 '18 at 13:31










  • @JohnDon do you mean $hat{f}$ is analytic in some nbhd?
    – mathworker21
    Dec 29 '18 at 13:34


















  • How does the result follow easily assuming $f$ is analytic in some nbhd?
    – mathworker21
    Dec 29 '18 at 13:01












  • @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).
    – John Don
    Dec 29 '18 at 13:11












  • 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
    – Michh
    Dec 29 '18 at 13:12






  • 1




    This is what you are looking for with the roles of $f$ and $hat{f}$ interchanged.
    – Michh
    Dec 29 '18 at 13:31










  • @JohnDon do you mean $hat{f}$ is analytic in some nbhd?
    – mathworker21
    Dec 29 '18 at 13:34
















How does the result follow easily assuming $f$ is analytic in some nbhd?
– mathworker21
Dec 29 '18 at 13:01






How does the result follow easily assuming $f$ is analytic in some nbhd?
– mathworker21
Dec 29 '18 at 13:01














@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).
– John Don
Dec 29 '18 at 13:11






@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).
– John Don
Dec 29 '18 at 13:11














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
– Michh
Dec 29 '18 at 13:12




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
– Michh
Dec 29 '18 at 13:12




1




1




This is what you are looking for with the roles of $f$ and $hat{f}$ interchanged.
– Michh
Dec 29 '18 at 13:31




This is what you are looking for with the roles of $f$ and $hat{f}$ interchanged.
– Michh
Dec 29 '18 at 13:31












@JohnDon do you mean $hat{f}$ is analytic in some nbhd?
– mathworker21
Dec 29 '18 at 13:34




@JohnDon do you mean $hat{f}$ is analytic in some nbhd?
– mathworker21
Dec 29 '18 at 13:34










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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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    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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      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
























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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.






        share|cite|improve this answer












        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 2 days ago









        DisintegratingByParts

        58.7k42579




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