Fourier transformation of bump functions
$begingroup$
Suppose we have $Omegasubset mathbb{R} $ open and we are looking at the space $ C_c^infty(Omega):={varphiin C^infty(Omega) | text{supp($varphi$) is a compact subset of $Omega$}} $. Now let $varphiin C_c^infty(Omega)$. Is the Fouriertransformation $(mathcal{F}varphi)(xi) $ in general an $L^1 $-function?
functional-analysis analysis measure-theory
edited Dec 12 '18 at 23:46
Bernard
121k740116
asked Dec 12 '18 at 22:32
KatakuriKatakuri
286
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add a comment |
$begingroup$
Suppose we have $Omegasubset mathbb{R} $ open and we are looking at the space $ C_c^infty(Omega):={varphiin C^infty(Omega) | text{supp($varphi$) is a compact subset of $Omega$}} $. Now let $varphiin C_c^infty(Omega)$. Is the Fouriertransformation $(mathcal{F}varphi)(xi) $ in general an $L^1 $-function?
functional-analysis analysis measure-theory
edited Dec 12 '18 at 23:46
Bernard
121k740116
asked Dec 12 '18 at 22:32
KatakuriKatakuri
286
$endgroup$
add a comment |
$begingroup$
Suppose we have $Omegasubset mathbb{R} $ open and we are looking at the space $ C_c^infty(Omega):={varphiin C^infty(Omega) | text{supp($varphi$) is a compact subset of $Omega$}} $. Now let $varphiin C_c^infty(Omega)$. Is the Fouriertransformation $(mathcal{F}varphi)(xi) $ in general an $L^1 $-function?
functional-analysis analysis measure-theory
edited Dec 12 '18 at 23:46
Bernard
121k740116
asked Dec 12 '18 at 22:32
KatakuriKatakuri
286
$endgroup$
Suppose we have $Omegasubset mathbb{R} $ open and we are looking at the space $ C_c^infty(Omega):={varphiin C^infty(Omega) | text{supp($varphi$) is a compact subset of $Omega$}} $. Now let $varphiin C_c^infty(Omega)$. Is the Fouriertransformation $(mathcal{F}varphi)(xi) $ in general an $L^1 $-function?
functional-analysis analysis measure-theory
functional-analysis analysis measure-theory
edited Dec 12 '18 at 23:46
Bernard
121k740116
asked Dec 12 '18 at 22:32
KatakuriKatakuri
286
edited Dec 12 '18 at 23:46
Bernard
121k740116
asked Dec 12 '18 at 22:32
KatakuriKatakuri
286
edited Dec 12 '18 at 23:46
Bernard
121k740116
edited Dec 12 '18 at 23:46
Bernard
121k740116
edited Dec 12 '18 at 23:46
Bernard
121k740116
121k740116
asked Dec 12 '18 at 22:32
KatakuriKatakuri
286
asked Dec 12 '18 at 22:32
KatakuriKatakuri
286
asked Dec 12 '18 at 22:32
KatakuriKatakuri
286
286
add a comment |
add a comment |
1 Answer
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$begingroup$
Any $f in C^{infty}_c(Omega)$ can be completed to be $f in C^{infty}_c(mathbb R)$ by just setting it to zero outside of $Omega$, so the Fourier transform makes sense. Then $C^{infty}_c(mathbb R)$ is a subset of a more general space of Schwartz functions. This class of functions has the nice property that the Fourier transform of a Schwartz function is a Schwartz function. In particular all Schwartz functions are $L^1$ so the result you want does hold.
If you want to prove this yourself, a hint is that the Fourier transform exchanges decay and regularity. So to prove $mathcal F f in L^1$, you need to use smoothness of $f$ to show $mathcal Ff$ has sufficient decay.
answered Dec 12 '18 at 22:43
bitesizebobitesizebo
1,50618
$endgroup$
$begingroup$
(here regularity means the derivatives are $L^1$)
$endgroup$
– reuns
Dec 13 '18 at 0:16
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
votes
$begingroup$
Any $f in C^{infty}_c(Omega)$ can be completed to be $f in C^{infty}_c(mathbb R)$ by just setting it to zero outside of $Omega$, so the Fourier transform makes sense. Then $C^{infty}_c(mathbb R)$ is a subset of a more general space of Schwartz functions. This class of functions has the nice property that the Fourier transform of a Schwartz function is a Schwartz function. In particular all Schwartz functions are $L^1$ so the result you want does hold.
If you want to prove this yourself, a hint is that the Fourier transform exchanges decay and regularity. So to prove $mathcal F f in L^1$, you need to use smoothness of $f$ to show $mathcal Ff$ has sufficient decay.
answered Dec 12 '18 at 22:43
bitesizebobitesizebo
1,50618
$endgroup$
$begingroup$
(here regularity means the derivatives are $L^1$)
$endgroup$
– reuns
Dec 13 '18 at 0:16
add a comment |
$begingroup$
Any $f in C^{infty}_c(Omega)$ can be completed to be $f in C^{infty}_c(mathbb R)$ by just setting it to zero outside of $Omega$, so the Fourier transform makes sense. Then $C^{infty}_c(mathbb R)$ is a subset of a more general space of Schwartz functions. This class of functions has the nice property that the Fourier transform of a Schwartz function is a Schwartz function. In particular all Schwartz functions are $L^1$ so the result you want does hold.
If you want to prove this yourself, a hint is that the Fourier transform exchanges decay and regularity. So to prove $mathcal F f in L^1$, you need to use smoothness of $f$ to show $mathcal Ff$ has sufficient decay.
answered Dec 12 '18 at 22:43
bitesizebobitesizebo
1,50618
$endgroup$
$begingroup$
(here regularity means the derivatives are $L^1$)
$endgroup$
– reuns
Dec 13 '18 at 0:16
add a comment |
$begingroup$
Any $f in C^{infty}_c(Omega)$ can be completed to be $f in C^{infty}_c(mathbb R)$ by just setting it to zero outside of $Omega$, so the Fourier transform makes sense. Then $C^{infty}_c(mathbb R)$ is a subset of a more general space of Schwartz functions. This class of functions has the nice property that the Fourier transform of a Schwartz function is a Schwartz function. In particular all Schwartz functions are $L^1$ so the result you want does hold.
If you want to prove this yourself, a hint is that the Fourier transform exchanges decay and regularity. So to prove $mathcal F f in L^1$, you need to use smoothness of $f$ to show $mathcal Ff$ has sufficient decay.
answered Dec 12 '18 at 22:43
bitesizebobitesizebo
1,50618
$endgroup$
Any $f in C^{infty}_c(Omega)$ can be completed to be $f in C^{infty}_c(mathbb R)$ by just setting it to zero outside of $Omega$, so the Fourier transform makes sense. Then $C^{infty}_c(mathbb R)$ is a subset of a more general space of Schwartz functions. This class of functions has the nice property that the Fourier transform of a Schwartz function is a Schwartz function. In particular all Schwartz functions are $L^1$ so the result you want does hold.
If you want to prove this yourself, a hint is that the Fourier transform exchanges decay and regularity. So to prove $mathcal F f in L^1$, you need to use smoothness of $f$ to show $mathcal Ff$ has sufficient decay.
answered Dec 12 '18 at 22:43
bitesizebobitesizebo
1,50618
edited Dec 12 '18 at 23:06
answered Dec 12 '18 at 22:43
bitesizebobitesizebo
1,50618
answered Dec 12 '18 at 22:43
bitesizebobitesizebo
1,50618
answered Dec 12 '18 at 22:43
bitesizebobitesizebo
1,50618
1,50618
$begingroup$
(here regularity means the derivatives are $L^1$)
$endgroup$
– reuns
Dec 13 '18 at 0:16
add a comment |
$begingroup$
(here regularity means the derivatives are $L^1$)
$endgroup$
– reuns
Dec 13 '18 at 0:16
$begingroup$
(here regularity means the derivatives are $L^1$)
$endgroup$
– reuns
Dec 13 '18 at 0:16
$begingroup$
(here regularity means the derivatives are $L^1$)
$endgroup$
– reuns
Dec 13 '18 at 0:16
add a comment |
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